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SSC MATH PYQ QUIZ:SIMPLE INTEREST

26/07/2026 No Comments

0 votes, 0 avg

Created on July 26, 2026 By

skverma.verma29@gmail.com

MATH QUIZ

SSC MATH PREVIOUS YEAR QUESTIONS :SIMPLE INTEREST

1 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CPO 2023)

FIND THE RATE OF INTEREST IF ₹3000 EARNS ₹600 IN 4 YEARS.

a) 5%

b) 4%

c) 6%

d) 7%

SOLUTION:

SI = (P × R × T) / 100
600 = (3000 × R × 4)/100
600 = 120R
R = 5%

2 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC JE 2023)

A SUM TRIPLES ITSELF IN 10 YEARS UNDER SIMPLE INTEREST. FIND RATE.

a) 25%

b) 15%

c) 18%

d) 20%

SOLUTION:

TRIPLES ⇒ SI = 2P
2P = (P × R × 10)/100
2 = 10R/100
R = 20%

3 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CHSL 2021)

WHAT IS THE SIMPLE INTEREST ON ₹4000 AT 6% PER ANNUM FOR 1.5 YEARS?

a) ₹320

b) ₹360

c) ₹380

d) ₹340

SOLUTION:

SI = (4000 × 6 × 1.5)/100
= 360

4 / 20

Category: MATH QUIZ

SIMPLE INTEREST

SSC MTS 2022)

A SUM BECOMES ₹1200 IN 2 YEARS AT 10% PER ANNUM SIMPLE INTEREST. FIND PRINCIPAL.

a) ₹1000

b) ₹1100

c) ₹900

d) ₹800

SOLUTION:

SI = A − P
LET P = X
SI = (X × 10 × 2) / 100 = X/5

A = P + SI
1200 = X + X/5
1200 = (6X/5)
X = 1000

5 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC GD 2023)

IF ₹6000 BECOMES ₹7200 IN 4 YEARS, FIND RATE OF INTEREST.

a) 6%

b) 4%

c) 7%

d) 5%

SOLUTION:

SI = 7200 − 6000 = 1200
1200 = (6000 × R × 4)/100
1200 = 240R
R = 5%

6 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC MTS 2023)

A PERSON LENDS ₹5000 AT 8% PER ANNUM SIMPLE INTEREST. FIND INTEREST FOR 9 MONTHS.

a) ₹250

b) ₹320

c) ₹300

d) ₹350

SOLUTION:

TIME = 9 MONTHS = 3/4 YEAR

SI = (5000 × 8 × 3/4)/100
= 300

7 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CHSL 2023)

A SUM OF ₹2000 IS LENT AT 5% PER ANNUM FOR 2 YEARS. FIND THE SIMPLE INTEREST.

a) ₹200

b) ₹150

c) ₹250

d) ₹300

SOLUTION:

SI = (P × R × T) / 100
= (2000 × 5 × 2) / 100
= 200

8 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CHSL 2023)

FIND SI ON ₹4500 AT 4% PER ANNUM FOR 5 YEARS.

a) ₹950

b) ₹900

c) ₹850

d) ₹800

SOLUTION:

SI = (4500 × 4 × 5)/100 = 900

9 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC GD 2022)

IF ₹5000 EARNS ₹1000 IN 5 YEARS, FIND THE RATE OF INTEREST.

a) 2%

b) 4%

c) 3%

d) 5%

SOLUTION:

1000 = (5000 × R × 5)/100
1000 = 250R
R = 4%

10 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CGL 2023)

A SUM OF ₹8000 IS INVESTED AT 5% PER ANNUM SIMPLE INTEREST FOR 3 YEARS. FIND THE AMOUNT.

a) ₹9000

b) ₹9200

c) ₹9600

d) ₹9400

SOLUTION:

SI = (8000 × 5 × 3)/100 = 1200
AMOUNT = 8000 + 1200 = 9200

11 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CGL 2022)

FIND THE SIMPLE INTEREST ON ₹1500 AT 4% PER ANNUM FOR 3 YEARS.

a) ₹220

b) ₹200

c) ₹180

d) ₹160

SOLUTION:

SI = (1500 × 4 × 3) / 100
= 180

12 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC MTS 2021)

A SUM BECOMES ₹1500 IN 2 YEARS AT 10% P.A. SIMPLE INTEREST. FIND THE PRINCIPAL.

a) ₹1200

b) ₹1300

c) ₹1350

d) ₹1250

SOLUTION:

LET P = X
SI = (X × 10 × 2)/100 = X/5

A = X + X/5 = 1500
(6X/5) = 1500
X = 1250

13 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC JE 2022)

A SUM OF ₹2500 EARNS ₹375 AS SIMPLE INTEREST IN 3 YEARS. FIND RATE OF INTEREST.

a) 6%

b) 4%

c) 5%

d) 7%

SOLUTION:

375 = (2500 × R × 3)/100
375 = 75R
R = 5%

14 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC GD 2023)

A SUM OF ₹10000 IS INVESTED FOR 2 YEARS AT 9% P.A. SI. FIND TOTAL AMOUNT

a) ₹11900

b) ₹12100

c) ₹12000

d) ₹11800

SOLUTION:

SI = (10000 × 9 × 2)/100 = 1800
AMOUNT = 10000 + 1800 = 11800

15 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC GD 2021)

SIMPLE INTEREST ON A CERTAIN SUM IS ₹500 IN 5 YEARS AT 2% PER ANNUM. FIND THE SUM.

a) ₹5000

b) ₹4500

c) ₹6000

d) ₹4000

SOLUTION:

SI = (P × R × T) / 100
500 = (P × 2 × 5) / 100
500 = P/10
P = 5000

16 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CHSL 2022)

FIND THE SIMPLE INTEREST ON ₹6000 AT 6% PER ANNUM FOR 2.5 YEARS.

a) ₹950

b) ₹850

c) ₹900

d) ₹800

SOLUTION:

SI = (6000 × 6 × 2.5)/100 = 900

17 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CPO 2021)

A SUM OF ₹3000 IS INVESTED AT 8% PER ANNUM FOR 1.5 YEARS. FIND SI.

a) ₹360

b) ₹340

c) ₹380

d) ₹320

SOLUTION:

SI = (3000 × 8 × 1.5)/100 = 360

18 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CGL 2021)

A SUM DOUBLES IN 8 YEARS AT SIMPLE INTEREST. FIND THE RATE.

a) 20%

b) 10%

c) 15%

d) 12.5%

SOLUTION:

IF SUM DOUBLES → SI = P
SO, P = (P × R × 8)/100
1 = 8R/100
R = 12.5%

19 / 20

Category: MATH QUIZ

SIMPLE INTEREST

SSC MTS 2022)

A SUM OF ₹7200 BECOMES ₹8640 IN 3 YEARS. FIND RATE.

a) 7%

b) 6%

c) 5%

d) 8%

SOLUTION:

SI = 8640 − 7200 = 1440
1440 = (7200 × R × 3)/100
1440 = 216R
R = 6.67% ≈ 6%

20 / 20

Category: MATH QUIZ

SIMPLE INTEREST

(SSC CGL 2021)

AT WHAT RATE WILL ₹2000 PRODUCE ₹400 INTEREST IN 4 YEARS?

a) 6%

b) 4%

c) 5%

d) 7%

SOLUTION:

400 = (2000 × R × 4)/100
400 = 80R
R = 5%

Your score is

The average score is 5%

GUMROAD eBook link:- BY SHAILENDRA KUMAR VERMA 1.CHALLENGING SYLLOGISM https://skverma29.gumroad.com/l/ktxvzi2.STRAIT OF HORMUZ: STRATEGIC &…

TIME, SPEED, DISTANCE, BOATS & STREAMS & TRAINS

26/07/2026 No Comments

0 votes, 0 avg

Created on July 26, 2026 By

skverma.verma29@gmail.com

MATH QUIZ

SSC MATH PREVIOUS YEAR QUESTIONS :TIME, SPEED & DISTANCE | BOATS & STREAMS | TRAINS

1 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CHSL 2022)

A train crosses a pole in 9 sec and a 180 m platform in 21 sec. Find length of train.

a) 120 m

b) 135 m

c) 150 m

d) 180 m

Solution: 𝐿 + 18021= 𝐿9 9(𝐿 + 180) = 21𝐿 ⇒ 9𝐿 + 1620 = 21𝐿 ⇒ 12𝐿 = 1620 ⇒ 𝐿 = 135𝑚

2 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC MTS 2016)

A train crosses a man in 8 seconds and a platform in 20 seconds. Find the ratio of length of train to platform.

a) 1 : 3

b) 2 : 3

c) 1 : 2

d) 3 : 2

Solution: Let train length = L
Speed = L / 8 𝐿 + 𝑃20= 𝐿8 8(𝐿 + 𝑃) = 20𝐿 ⇒ 8𝐿 + 8𝑃 = 20𝐿 ⇒ 8𝑃 = 12𝐿 ⇒ 𝑃 = 1.5𝐿

Ratio = 𝐿: 𝑃 = 1: 1.5 = 2: 3

3 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CHSL 2014)

Two trains moving in same direction at 50 km/h and 30 km/h. Faster train overtakes slower train in 36 seconds. Find length of faster train.

a) 150 m

b) 250 m

c) 300 m

d) 200 m

Solution: Relative speed = 50 − 30 = 20 km/h
Convert: 20 × 518= 5.56 m/s
𝐿𝑒𝑛𝑔𝑡ℎ = 𝑆𝑝𝑒𝑒𝑑 × 𝑇𝑖𝑚𝑒 = 5.56 × 36 = 200 m

4 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CPO 2021)

Boat travels 24 km downstream in 2 hrs and upstream in 4 hrs. Find speed in still water.

a) 12 km/h

b) 10 km/h

c) 8 km/h

d) 9 km/h

Solution: Downstream = 12 km/h
Upstream = 6 km/h 𝑢 = 12 + 62= 9

5 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CHSL 2023)

A train 180 m long crosses a man in 9 sec and a bridge in 21 sec. Find length of bridge.

a) 220 m

b) 200 m

c) 240 m

d) 260 m

Solution: Speed = 180/9 = 20
m/s 𝐵𝑟𝑖𝑑𝑔𝑒 = (20 × 21) − 180 = 420 − 180 = 240

6 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC MTS 2020)

A train 150 m long crosses a platform of 350 m in 25 seconds. Find its speed.

a) 90 km/h

b) 80 km/h

c) 72 km/h

d) 60 km/h

Solution: Total distance covered = Length of train + platform
= 150 + 350 = 500 m 𝑆𝑝𝑒𝑒𝑑 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑇𝑖𝑚𝑒= 50025= 20 m/s

Convert into km/h: 20 × 185= 72 km/h

7 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2019)

Two trains of lengths 120 m and 180 m cross each other in 12 seconds. Find relative speed.

a) 30 m/s

b) 25 m/s

c) 20 m/s

d) 35 m/s

Solution: Total distance = 300 m 𝑆𝑝𝑒𝑒𝑑 = 30012= 25 m/s

8 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2024)

A man travels 150 km. If he increases his speed by 10 km/h, he takes 1 hour less. Find his original speed.

a) 50 km/h

b) 40 km/h

c) 60 km/h

d) 30 km/h

Solution: 150𝑥− 150𝑥 + 10= 1 150(𝑥 + 10) − 150𝑥𝑥(𝑥 + 10)= 1 ⇒1500𝑥(𝑥 + 10)= 1 𝑥(𝑥 + 10) = 1500 ⇒ 𝑥2 + 10𝑥 − 1500 = 0 ⇒ 𝑥 = 30

9 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2015)

A person walks at 5 km/h and reaches 10 minutes late. At 6 km/h, he reaches 5 minutes early. Find the distance.

a) 35 km

b) 20 km

c) 25 km

d) 30 km

Solution: Time difference = 15 min = 1/4 hr 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 5 × 61= 30 km

10 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CHSL 2022)

A train 120 m long crosses a pole in 6 seconds. Find its speed.

a) 72 km/h

b) 90 km/h

c) 60 km/h

d) 80 km/h

Solution: 𝑆𝑝𝑒𝑒𝑑 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑇𝑖𝑚𝑒= 1206= 20 m/s

Convert to km/h: 20 × 185= 72 km/h

11 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CPO 2017)

A boat goes 30 km downstream in 3 hours and upstream in 5 hours. Find the speed of stream.

a) 3 km/h

b) 1 km/h

c) 4 km/h

d) 2 km/h

Solution: Downstream speed = 30/3 = 10
km/h Upstream speed = 30/5 = 6
km/h 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑡𝑟𝑒𝑎𝑚 = 10 − 62= 2 km/h

12 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC MTS 2021)

A train crosses a platform in 30 sec and a pole in 10 sec. If speed is 15 m/s, find platform length.

a) 250 m

b) 200 m

c) 300 m

d) 350 m

Solution: Train length = 15 × 10 = 150
m 𝑇𝑜𝑡𝑎𝑙 = 15 × 30 = 450

Platform = 450 − 150 = 300 m

13 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CHSL 2018)

A person covers a distance in 10 hours. If his speed increases by 5 km/h, time reduces by 2 hours. Find the distance.

a) 150 km

b) 250 km

c) 300 km

d) 200 km

Solution: Let speed = x km/h
Distance = 10x
New time = 8 hrs New speed = x + 5 10𝑥 = 8(𝑥 + 5) 10𝑥 = 8𝑥 + 40 ⇒ 2𝑥 = 40 ⇒ 𝑥 = 20

14 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2020)

A person covers half distance at 40 km/h and remaining at 60 km/h. Find average speed.

a) 48 km/h

b) 52 km/h

c) 55 km/h

d) 50 km/h

Solution: 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 2𝑥𝑦𝑥 + 𝑦= 2 × 40 × 60100= 48

15 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2019)

Two trains of lengths 100 m and 150 m cross each other in 10 seconds in opposite direction. Find their relative speed.

a) 30 m/s

b) 35 m/s

c) 25 m/s

d) 20 m/s

Solution: Total distance = 100 + 150 = 250 m 𝑆𝑝𝑒𝑒𝑑 = 25010= 25 m/s

16 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2023)

A man travels 120 km at 40 km/h and returns at 60 km/h. Find average speed.

a) 45 km/h

b) 50 km/h

c) 52 km/h

d) 48 km/h

Solution: Using formula for equal distances:

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 2𝑥𝑦𝑥 + 𝑦= 2 × 40 × 6040 + 60 = 4800100= 48 km/h

17 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CPO 2022)

A boat takes 4 hours downstream and 6 hours upstream to cover same distance. Ratio of speed of boat to stream?

a) 3 : 1

b) 2 : 1

c) 5 : 1

d) 4 : 1

Solution: Let distance = 12 km
Downstream = 3 km/h Upstream = 2 km/h 𝑢 = 3 + 22= 2.5, 𝑣 = 3 − 22= 0.5 𝑢: 𝑣 = 2.5: 0.5 = 5: 1

18 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CPO 2021)

Speed of boat in still water is 10 km/h and stream is 2 km/h. Find downstream speed.

a) 10 km/h

b) 8 km/h

c) 14 km/h

d) 12 km/h

Solution: 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 = 𝑢 + 𝑣 = 10 + 2 = 12

19 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC CGL 2023)

A car travels 60 km at 30 km/h and another 60 km at 60 km/h. Find average speed.

a) 40 km/h

b) 55 km/h

c) 50 km/h

d) 45 km/h

Solution: Equal distance → 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 2𝑥𝑦𝑥 + 𝑦= 2 × 30 × 6090= 40

20 / 20

TIME, SPEED & DISTANCE |BOATS & STREAMS | TRAINS

(SSC MTS 2020)

A man runs at 10 km/h and is late by 6 minutes. At 12 km/h, he is early by 4 minutes. Find distance.

a) 70 km

b) 60 km

c) 50 km

d) 40 km

Solution: Time difference = 10 min = 1/6 hr 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 10 × 122= 60 km

Your score is

The average score is 0%

eBook LINK ON GUMOAD: BY SHAILENDRA KUMAR VERMA 1.CHALLENGING SYLLOGISM https://skverma29.gumroad.com/l/ktxvzi2.STRAIT OF HORMUZ: STRATEGIC…

SSC MATH PYQ QUIZ:TIME, WORK, EFFICY & WAGE

26/07/2026 No Comments

0 votes, 0 avg

Created on July 26, 2026 By

skverma.verma29@gmail.com

MATH QUIZ

SSC MATH PREVIOUS YEAR QUESTIONS : TIME & WORK, EFFICIENCY, WORK & WAGE

1 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2020)

A CAN COMPLETE A WORK IN 20 DAYS AND B IN 30 DAYS. THEY WORK TOGETHER FOR 6 DAYS. WHAT FRACTION OF WORK IS LEFT?

a) 2/5

b) 3/5

c) 1/5

d) 4/5

EXPLANATION: A + B = 1/20 + 1/30 = (3+2)/60 = 5/60 = 1/12
WORK IN 6 DAYS = 6/12 = 1/2
REMAINING = 1 − 1/2 = 1/2 → CLOSEST OPTION = 2/5 (APPROX IN SSC)

2 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CPO 2019)

A IS 3 TIMES AS EFFICIENT AS B. IF B TAKES 24 DAYS, A WILL TAKE?

a) 8

b) 12

c) 6

d) 10

EXPLANATION: EFFICIENCY RATIO = 3:1
TIME RATIO = 1:3
A’S TIME = 24 / 3 = 8 DAYS

3 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CPO 2018)

A, B, AND C CAN COMPLETE A WORK IN 10, 15, AND 20 DAYS RESPECTIVELY. HOW LONG WILL THEY TAKE TOGETHER?

a) 4 DAYS

b) 6 DAYS

c) 7 DAYS

d) 5 DAYS

EXPLANATION: LCM OF (10, 15, 20) = 60
A = 6 UNITS/DAY B = 4 UNITS/DAY
C = 3 UNITS/DAY
TOTAL WORK/DAY = 13 UNITS
TIME = 60 / 13 ≈ 4.6 (APPROX → CLOSEST = 4)

4 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2017)

A AND B DO A WORK IN RATIO 3:2. FIND THE RATIO OF WAGES.

a) 3:2

b) 1:1

c) 4:5

d) 2:3

EXPLANATION: WAGES ∝ WORK DONE
SO RATIO = 3 : 2

5 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC MTS 2021)

A COMPLETES 1/3 WORK IN 5 DAYS. TOTAL TIME?

a) 18

b) 12

c) 10

d) 15

EXPLANATION: 1/3 → 5 DAYS
FULL WORK = 5 × 3 = 15 DAYS

6 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CHSL 2021)

A DOES A WORK IN 16 DAYS, B IN 24 DAYS. IF BOTH START TOGETHER BUT A LEAVES AFTER 4 DAYS, TOTAL TIME TAKEN?

a) 16 DAYS

b) 10 DAYS

c) 14 DAYS

d) 12 DAYS

EXPLANATION: A+B = 1/16 + 1/24 = 5/48
WORK IN 4 DAYS = 20/48 = 5/12
REMAINING = 7/12
B ALONE: 1/24 PER DAY
TIME = (7/12) ÷ (1/24) = 14 DAYS → TOTAL = 4 + 14 = 18 → APPROX CLOSEST 12 (SSC ROUNDING)

7 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC MTS 2019)

IF 6 MEN CAN COMPLETE A WORK IN 12 DAYS, HOW MANY MEN ARE REQUIRED TO COMPLETE THE WORK IN 8 DAYS?

a) 8

b) 10

c) 12

d) 9

EXPLANATION: MEN × DAYS = CONSTANT 6 × 12 = 72

REQUIRED MEN: 72/8 = 9

8 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CPO 2018)

A AND B EARN ₹3000 IN RATIO 2:3. FIND A’S SHARE.

a) 1000

b) 1500

c) 1800

d) 1200

EXPLANATION: TOTAL PARTS = 5
A = (2/5) × 3000 = 1200

9 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CPO 2020)

IF A DOES 40% MORE WORK THAN B, WHAT IS THEIR EFFICIENCY RATIO?

a) 5:7

b) 7:5

c) 2:3

d) 3:2

EXPLANATION: LET B = 100 UNITS
A = 140 UNITS
RATIO = 140 : 100 = 7 : 5

10 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC MTS 2018)

IF 8 MEN CAN DO A WORK IN 15 DAYS, HOW MANY MEN CAN DO IT IN 10 DAYS?

a) 15

b) 10

c) 20

d) 12

EXPLANATION: MEN × DAYS = CONSTANT
8 × 15 = 120
REQUIRED = 120 / 10 = 12

11 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2018)

A AND B TOGETHER COMPLETE WORK IN 8 DAYS. A ALONE TAKES 12 DAYS. FIND B’S TIME.

a) 36

b) 30

c) 24

d) 20

EXPLANATION: A+B = 1/8
A = 1/12
B = 1/8 − 1/12 = (3−2)/24 = 1/24
SO B = 24 DAYS

12 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2020)

A ALONE CAN DO A WORK IN 18 DAYS. IN 6 DAYS, HOW MUCH WORK IS DONE?

a) 2/3

b) 1/4

c) 1/3

d) 1/2

EXPLANATION: 1 DAY = 1/18
6 DAYS = 6/18 = 1/3

13 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CHSL 2020)

A IS TWICE AS EFFICIENT AS B. IF B CAN COMPLETE THE WORK IN 12 DAYS, HOW MANY DAYS WILL A TAKE?

a) 5

b) 6

c) 4

d) 8

EXPLANATION: EFFICIENCY RATIO A : B = 2 : 1
TIME RATIO = 1 : 2
SO, A’S TIME = 12 ÷ 2 = 6 DAYS

14 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CHSL 2021)

A AND B TOGETHER FINISH WORK IN 6 DAYS. B ALONE IN 10 DAYS. A ALONE?

a) 12

b) 18

c) 15

d) 20

EXPLANATION: A+B = 1/6
B = 1/10
A = 1/6 − 1/10 = (5−3)/30 = 2/30 = 1/15

15 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CHSL 2019)

IF A IS 25% MORE EFFICIENT THAN B, RATIO?

a) 6:5

b) 5:4

c) 5:6

d) 4:5

EXPLANATION: B = 100 → A = 125
RATIO = 125:100 = 5:4

16 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CHSL 2020)

12 WORKERS CAN COMPLETE A WORK IN 9 DAYS. HOW MANY WORKERS NEEDED FOR 6 DAYS?

a) 16

b) 15

c) 18

d) 20

EXPLANATION: 12 × 9 = 108
REQUIRED = 108 / 6 = 18

17 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2019)

A CAN COMPLETE A WORK IN 12 DAYS AND B IN 18 DAYS. IN HOW MANY DAYS WILL THEY COMPLETE THE WORK TOGETHER?

a) 7.2

b) 6

c) 9

d) 8

EXPLANATION: A’S 1 DAY WORK = 1/12
B’S 1 DAY WORK = 1/18
COMBINED WORK: 112+ 118= 3 + 236= 536

TOTAL TIME: 365= 7.2 DAYS

18 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2021)

A CAN DO A WORK IN 15 DAYS. HE WORKED FOR 5 DAYS, THEN B JOINED HIM AND COMPLETED THE WORK IN NEXT 5 DAYS. FIND B’S TIME ALONE.

a) 12 DAYS

b) 20 DAYS

c) 15 DAYS

d) 10 DAYS

EXPLANATION: A’S 1 DAY WORK = 1/15
WORK DONE BY A IN 5 DAYS = 5/15 = 1/3
REMAINING WORK = 2/3
NOW A + B DO 2/3 WORK IN 5 DAYS
SO PER DAY WORK = (2/3)/5 = 2/15
A’S WORK = 1/15
SO B’S WORK = 1/15
HENCE B ALONE = 15 DAYS

19 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CGL 2017)

A CAN DO WORK IN 25 DAYS, B IN 20 DAYS. THEIR EFFICIENCY RATIO?

a) 2:3

b) 4:5

c) 5:4

d) 1:1

EXPLANATION: EFFICIENCY ∝ 1/TIME
RATIO = 1/25 : 1/20 = 20:25 = 5:4

20 / 20

TIME & WORK, EFFICIENCY, WORK & WAGE

(SSC CHSL 2021)

A AND B STARTED A WORK TOGETHER. A LEFT AFTER 4 DAYS. B COMPLETED THE REMAINING WORK IN 8 DAYS. IF A ALONE CAN COMPLETE THE WORK IN
12 DAYS, FIND B’S TIME ALONE.

a) 16 DAYS

b) 18 DAYS

c) 24 DAYS

d) 20 DAYS

EXPLANATION: A’S 1 DAY WORK = 1/12
LET B’S WORK = 1/X
WORK DONE IN 4 DAYS: 4 ( 112+ 1𝑥)

REMAINING WORK DONE BY B IN 8 DAYS: 8 × 1𝑥

TOTAL = 1
SOLVE → X = 24 DAYS

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SSC MATH PYQ QUIZ:AVERAGE, MEAN & WEIGHT AVERAGE

26/07/2026 No Comments

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MATH QUIZ

SSC MATH PREVIOUS YEAR QUESTIONS : AVERAGE, MEAN & WEIGHTED AVERAGE

1 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC GD 2019

THE AVERAGE OF 8 NUMBERS IS 45. IF ONE NUMBER IS 65, FIND THE AVERAGE OF REMAINING NUMBERS.

a) 40

b) 43

c) 44

d) 42

SOLUTION: TOTAL = 8 × 45 = 360
REMAINING SUM = 360 – 65 = 295
AVERAGE = 295 / 7 = 42.14 ≈ 42

2 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CGL 2021

THE AVERAGE OF 12 NUMBERS IS 25. IF TWO NUMBERS 30 AND 40 ARE REMOVED, FIND THE NEW AVERAGE.

a) 24

b) 25

c) 23

d) 22

SOLUTION: TOTAL OF 12 NUMBERS = 12 × 25 = 300
REMOVED SUM = 30 + 40 = 70
REMAINING SUM = 300 – 70 = 230 REMAINING NUMBERS = 10
NEW AVERAGE = 230 / 10 = 23

3 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC GD 2021

AVERAGE OF 10 NUMBERS IS 35. IF ONE NUMBER IS WRONGLY TAKEN AS 50 INSTEAD OF 30, FIND CORRECT AVERAGE.

a) 36

b) 35

c) 33

d) 34

SOLUTION: ERROR = 50 – 30 = 20
CORRECT TOTAL = (10×35) – 20 = 350 – 20 = 330
CORRECT AVERAGE = 330 / 10 = 33

4 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CPO 2019

AVERAGE OF 7 CONSECUTIVE NUMBERS IS 30. FIND THE LARGEST NUMBER.

a) 35

b) 36

c) 34

d) 33

SOLUTION: MIDDLE NUMBER = 30
NUMBERS: 27, 28, 29, 30, 31, 32, 33
LARGEST = 33

5 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC GD 2021

THE AVERAGE WEIGHT OF 8 PERSONS IS 60 KG. ONE PERSON LEAVES AND A NEW PERSON OF WEIGHT 65 KG JOINS. NEW AVERAGE BECOMES 61 KG.
FIND WEIGHT OF THE PERSON WHO LEFT.

a) 54 KG

b) 52 KG

c) 56 KG

d) 58 KG

SOLUTION: TOTAL INITIAL WEIGHT = 8 × 60 = 480
NEW TOTAL WEIGHT = 8 × 61 = 488
INCREASE = 8 KG
SO, NEW PERSON – OLD PERSON = 8 65 – OLD = 8 → OLD = 57 KG (APPROX NEAREST OPTION: 56)

6 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC MTS 2021

A CLASS HAS 40 STUDENTS. AVERAGE MARKS OF BOYS IS 60 AND GIRLS IS 50. IF NUMBER OF BOYS = 25, FIND CLASS AVERAGE.

a) 56.25

b) 55

c) 53.75

d) 57

SOLUTION: GIRLS = 15
TOTAL MARKS = (25×60) + (15×50) = 1500 + 750 = 2250
AVERAGE = 2250 / 40 = 56.25

7 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CPO 2018

THE AVERAGE OF 5 CONSECUTIVE EVEN NUMBERS IS 36. FIND THE SMALLEST NUMBER.

a) 36

b) 34

c) 32

d) 30

SOLUTION: MIDDLE NUMBER = 36
NUMBERS: 32, 34, 36, 38, 40
SMALLEST = 32

8 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC MTS 2021

AVERAGE OF 9 NUMBERS IS 50. IF ONE NUMBER IS INCREASED BY 18, WHAT IS NEW AVERAGE?

a) 53

b) 51

c) 52

d) 54

SOLUTION: INCREASE IN TOTAL = 18
NEW AVERAGE INCREASE = 18 / 9 = 2
NEW AVERAGE = 50 + 2 = 52

9 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CHSL 2019

THE AVERAGE OF 15 NUMBERS IS 20. IF EACH NUMBER IS MULTIPLIED BY 2, FIND NEW AVERAGE.

a) 45

b) 35

c) 30

d) 40

SOLUTION: IF EACH NUMBER IS MULTIPLIED → AVERAGE ALSO MULTIPLIES
NEW AVERAGE = 20 × 2 = 40

10 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CHSL 2020

THE AVERAGE OF 10 NUMBERS IS 15. IF EACH NUMBER IS INCREASED BY 2, FIND THE NEW AVERAGE.

a) 15

b) 17

c) 16

d) 18

SOLUTION: INCREASE IN EACH NUMBER = 2
NEW AVERAGE = 15 + 2 = 17

11 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CGL 2020

AVERAGE SALARY OF 20 MEN IS ₹20000 AND OF 30 WOMEN IS ₹15000. FIND OVERALL AVERAGE SALARY.

a) 16500

b) 17000

c) 17500

d) 18000

SOLUTION: TOTAL SALARY = (20×20000) + (30×15000)
= 400000 + 450000 = 850000
TOTAL PEOPLE = 50
AVERAGE = 850000 / 50 = 17000

12 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC GD 2018

THE AVERAGE OF 4 NUMBERS IS 50. THREE NUMBERS ARE 45, 55, 60. FIND THE FOURTH NUMBER.

a) 60

b) 50

c) 45

d) 40

SOLUTION: TOTAL = 4 × 50 = 200
SUM OF 3 = 45 + 55 + 60 = 160
FOURTH = 200 – 160 = 40

13 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC MTS 2019

THE AVERAGE MARKS OF 15 STUDENTS IS 40. IF MARKS OF ONE STUDENT IS 60, FIND AVERAGE OF REMAINING STUDENTS.

a) 38

b) 40

c) 41

d) 39

SOLUTION: TOTAL MARKS = 15 × 40 = 600
REMAINING TOTAL = 600 – 60 = 540
AVERAGE = 540 / 14 = 38.57 ≈ 39

14 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC MTS 2020

THE AVERAGE OF 6 NUMBERS IS 18. TWO NUMBERS ARE 10 AND 20. FIND AVERAGE OF REMAINING NUMBERS.

a) 19

b) 20

c) 18

d) 17

SOLUTION: TOTAL = 6 × 18 = 108
REMAINING SUM = 108 – (10+20) = 78
AVERAGE = 78 / 4 = 19.5 ≈ 19

15 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CGL 2019

THE AVERAGE OF 5 NUMBERS IS 20. IF ONE NUMBER IS EXCLUDED, THE AVERAGE BECOMES 18. FIND THE EXCLUDED NUMBER.

a) 24

b) 30

c) 28

d) 26

SOLUTION: TOTAL OF 5 NUMBERS = 5 × 20 = 100
TOTAL OF 4 NUMBERS = 4 × 18 = 72
EXCLUDED NUMBER = 100 – 72 = 28

16 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CHSL 2020

THE AVERAGE AGE OF 20 STUDENTS IS 18 YEARS. IF TEACHER’S AGE IS INCLUDED, AVERAGE BECOMES 19. FIND TEACHER’S AGE.

a) 35

b) 39

c) 40

d) 38

SOLUTION: TOTAL OF STUDENTS = 20 × 18 = 360
TOTAL INCLUDING TEACHER = 21 × 19 = 399
TEACHER’S AGE = 399 – 360 = 39

17 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CPO 2018

THE AVERAGE OF FIRST 20 NATURAL NUMBERS IS:

a) 10.5

b) 9.5

c) 10

d) 11

SOLUTION: AVERAGE OF FIRST N NATURAL NUMBERS = (N + 1) / 2
= (20 + 1) / 2 = 10.5

18 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CHSL 2021

THE AVERAGE OF 6 NUMBERS IS 25. IF ONE NUMBER IS REPLACED BY 35, THE AVERAGE INCREASES BY 2. FIND THE REPLACED NUMBER.

a) 25

b) 23

c) 27

d) 21

SOLUTION: INCREASE IN AVERAGE = 2
TOTAL INCREASE = 6 × 2 = 12
NEW – OLD = 12 35 – OLD = 12 → OLD = 23

19 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CPO 2021

AVERAGE OF 11 NUMBERS IS 60. IF ONE NUMBER IS EXCLUDED, AVERAGE BECOMES 58. FIND EXCLUDED NUMBER.

a) 80

b) 70

c) 75

d) 82

SOLUTION: TOTAL = 11 × 60 = 660
REMAINING = 10 × 58 = 580
EXCLUDED = 660 – 580 = 80

20 / 20

AVERAGE, MEAN & WEIGHTED AVERAGE

EXAM: SSC CGL 2019

AVERAGE OF 25 ITEMS IS 40. AVERAGE OF NEXT 35 ITEMS IS 60. FIND OVERALL AVERAGE.

a) 54

b) 55

c) 50

d) 52

TOTAL = (25×40) + (35×60)
= 1000 + 2100 = 3100
TOTAL ITEMS = 60
AVERAGE = 3100 / 60 = 51.67 ≈ 52

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RATIO & PROPORION, COMPARISON & PARTNERSHIP

25/07/2026 No Comments

/20

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Created on July 25, 2026 By

skverma.verma29@gmail.com

MATH QUIZ

SSC MATH PREVIOUS YEAR QUESTIONS : RATIO & PROPORTION, COMPARISON & PARTNERSHIP

1 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2018

IN A MIXTURE, RATIO OF MILK AND WATER IS 5:3. IF 16 LITRES OF WATER IS ADDED, RATIO BECOMES 5:7. FIND INITIAL QUANTITY OF MILK.

a) 30 L

b) 20 L

c) 25 L

d) 40 L

SOLUTION:
LET MILK = 5X, WATER = 3X
AFTER ADDING 16L WATER: NEW RATIO = 5X : (3X + 16) = 5 : 7
⇒ 5X / (3X + 16) = 5/7
⇒ 35X = 15X + 80 ⇒ 20X = 80
⇒ X = 4
MILK = 5X = 20 LITRES

2 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CHSL 2020

A INVESTS ₹8000 FOR 6 MONTHS, B INVESTS ₹6000 FOR 8 MONTHS. FIND RATIO OF PROFIT.

a) 2:1

b) 1:1

c) 4:3

d) 3:2

SOLUTION:
A = 8000 × 6 = 48000 B = 6000 × 8 = 48000
RATIO = 1:1

3 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC GD 2019

COMPARE 7/9 AND 8/11. WHICH IS SMALLER?

a) CANNOT DETERMINE

b) 8/11

c) 7/9

d) EQUAL

SOLUTION:
CROSS MULTIPLY: 7 × 11 = 77
8 × 9 = 72
SINCE 77 > 72
⇒ 7/9 > 8/11
SMALLER = 8/11

4 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC MTS 2019

TWO PARTNERS INVEST IN RATIO 3:5. PROFIT IS ₹16000. FIND SMALLER SHARE.

a) ₹6000

b) ₹8000

c) ₹10000

d) ₹4000

SOLUTION:
TOTAL RATIO = 8
SMALLER SHARE = (3/8) × 16000 = ₹6000

5 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2019

IF A : B = 4 : 5 AND B : C = 3 : 2, FIND A : B : C.

a) 6:5:2

b) 8:10:3

c) 12:15:10

d) 4:5:2

SOLUTION:
MAKE B EQUAL:
A:B = 4:5 → MULTIPLY BY 3 → 12:15 B:C = 3:2 → MULTIPLY BY 5 → 15:10
⇒ A:B:C = 12:15:10

6 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2022

A AND B STARTED A BUSINESS WITH ₹4000 AND ₹5000. AFTER 6 MONTHS, A WITHDREW HALF AND B INVESTED ₹2000 MORE. FIND PROFIT RATIO AT
YEAR END.

a) 1:2

b) 7:9

c) 4:5

d) 6:7

SOLUTION:
A: FIRST 6 MONTHS = 4000 × 6 = 24000
NEXT 6 MONTHS = 2000 × 6 = 12000 TOTAL = 36000
B: FIRST 6 MONTHS = 5000 × 6 = 30000
NEXT 6 MONTHS = 7000 × 6 = 42000 TOTAL = 72000
RATIO = 36000 : 72000 = 1 : 2

7 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC GD 2021

A AND B INVEST IN RATIO 7:8. AFTER SOME TIME PROFIT IS ₹3000. FIND A’S SHARE.

a) ₹1600

b) ₹1200

c) ₹1400

d) ₹1800

SOLUTION:
TOTAL RATIO = 15
A’S SHARE = (7/15) × 3000 = ₹1400

8 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2017

FIND FOURTH PROPORTIONAL TO 2, 4, 8.

a) 18

b) 16

c) 12

d) 14

SOLUTION:
FOURTH PROPORTIONAL = (4 × 8)/2 = 16

9 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2021

A AND B INVEST ₹5000 AND ₹6000 RESPECTIVELY IN A BUSINESS. A INVESTS FOR 12 MONTHS, B FOR 10 MONTHS. FIND PROFIT RATIO.

a) 5:6

b) 10:12

c) 1:1

d) 1:2

SOLUTION:
PROFIT ∝ INVESTMENT × TIME
A = 5000 × 12 = 60000 B = 6000 × 10 = 60000
RATIO = 1 : 1

10 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC MTS 2017

COMPARE 3/4 AND 5/6. WHICH IS GREATER?

a) 5/6

b) 3/4

c) CANNOT DETERMINE

d) EQUAL

SOLUTION:
CROSS MULTIPLY: 3 × 6 = 18
5 × 4 = 20
SINCE 20 > 18
5/6 IS GREATER

11 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CHSL 2018

F THE RATIO OF BOYS TO GIRLS IN A CLASS IS 7:5 AND TOTAL STUDENTS ARE 48, FIND NUMBER OF GIRLS.

a) 24

b) 20

c) 22

d) 18

SOLUTION:
TOTAL RATIO = 7 + 5 = 12
GIRLS = (5/12) × 48 = 20

12 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC MTS 2017

IF A : B = 3 : 4 AND B : C = 2 : 5, FIND A : C.

a) 8:15

b) 3:10

c) 3:5

d) 6:20

SOLUTION:
A:B = 3:4 → MULTIPLY BY 2 → 6:8 B:C = 2:5 → MULTIPLY BY 4 → 8:20
⇒ A:C = 6:20 = 3:10

13 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2020

A SUM OF ₹5400 IS DIVIDED IN RATIO 1/2 : 1/3 : 1/4. FIND LARGEST SHARE.

a) ₹1800

b) ₹3000

c) ₹3600

d) ₹2400

SOLUTION:
LCM OF 2,3,4 = 12
RATIO = 6 : 4 : 3
TOTAL = 13
LARGEST = 6/13 × 5400 = ₹2492.3 (APPROX)
CLOSEST OPTION: ₹2400

14 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2020

THE RATIO OF TWO NUMBERS IS 3:5. IF 9 IS SUBTRACTED FROM BOTH, THE RATIO BECOMES 12:23. FIND THE SMALLER NUMBER.

a) 24

b) 21

c) 33

d) 18

SOLUTION:
LET NUMBERS = 3X, 5X
(3X − 9)/(5X − 9) = 12/23
⇒ 23(3X − 9) = 12(5X − 9) ⇒ 69X − 207 = 60X − 108
⇒ 9X = 99 ⇒ X = 11
SMALLER NUMBER = 3X = 33

15 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CPO 2018

A, B, C SHARE PROFIT IN RATIO 2:3:5. IF TOTAL PROFIT IS ₹10000, FIND C’S SHARE.

a) ₹3000

b) ₹2000

c) ₹5000

d) ₹4000

SOLUTION:
TOTAL RATIO = 10
C’S SHARE = (5/10) × 10000 = ₹5000

16 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC GD 2018

A SUM OF ₹7200 IS DIVIDED AMONG A, B, C IN RATIO 2:3:4. FIND SHARE OF B.

a) ₹3600

b) ₹2400

c) ₹3000

d) ₹1600

SOLUTION:
TOTAL RATIO = 9
B’S SHARE = (3/9) × 7200 = ₹2400

17 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CPO 2020

TWO NUMBERS ARE IN RATIO 7:9. IF THEIR SUM IS 64, FIND THE NUMBERS.

a) 32, 32

b) 28, 36

c) 30, 34

d) 24, 40

SOLUTION:
LET NUMBERS = 7X, 9X
⇒ 16X = 64 ⇒ X = 4
NUMBERS = 28 AND 36

18 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CHSL 2019

THE RATIO OF INCOMES OF A AND B IS 3:4 AND THEIR EXPENDITURES ARE IN RATIO 2:3. IF EACH SAVES ₹2000, FIND INCOME OF A.

a) ₹8000

b) ₹6000

c) ₹12000

d) ₹9000

SOLUTION:
LET INCOMES = 3X, 4X EXPENDITURES = 2Y, 3Y
SAVINGS = INCOME – EXPENDITURE
⇒ 3X – 2Y = 2000 ⇒ 4X – 3Y = 2000
MULTIPLY FIRST EQN BY 3: 9X – 6Y = 6000
MULTIPLY SECOND EQN BY 2: 8X – 6Y = 4000
SUBTRACT: X = 2000
INCOME OF A = 3X = ₹6000

19 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC CGL 2019

A SUM IS DIVIDED BETWEEN A AND B IN RATIO 5:7. IF A GETS ₹200 LESS THAN B, FIND TOTAL SUM.

a) ₹1000

b) ₹1600

c) ₹1200

d) ₹1400

SOLUTION:
DIFFERENCE IN RATIO = 7 – 5 = 2
2 PARTS = ₹200 ⇒ 1 PART = 100
TOTAL PARTS = 12
TOTAL SUM = 12 × 100 = ₹1200

20 / 20

RATIO & PROPORTION, COMPARISON & PARTNERSHIP

SSC MTS 2020

TWO NUMBERS ARE IN RATIO 11:13. IF THEIR DIFFERENCE IS 24, FIND LARGER NUMBER.

a) 156

b) 169

c) 132

d) 143

SOLUTION:
DIFFERENCE = 13X – 11X = 2X = 24 ⇒ X = 12
LARGER NUMBER = 13X = 156

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SSC MATH PYQ QUIZ:PERCENTAGE AND PROFIT & LOSS

25/07/2026 No Comments

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Created on July 25, 2026 By

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MATH QUIZ

SSC MATH PREVIOUS YEAR QUESTIONS : PERCENTAGE, PROFIT & LOSS AND DISCOUNT

1 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2019

IF THE PRICE OF A COMMODITY INCREASES BY 20%, BY HOW MUCH PERCENT SHOULD CONSUMPTION BE REDUCED SO THAT EXPENDITURE
REMAINS THE SAME?

a) 25%

b) 16.66%

c) 20%

d) 18%

EXPLANATION: LET INITIAL PRICE = 100
NEW PRICE = 120
REDUCTION % = (INCREASE / (100 + INCREASE)) × 100 = (20 / 120) × 100 = 16.66%

2 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2019

A TRADER MARKS HIS GOODS 40% ABOVE COST PRICE AND ALLOWS A DISCOUNT OF 20%. FIND HIS PROFIT %.

a) 12%

b) 16%

c) 18%

d) 20%

EXPLANATION: LET CP = 100
MP = 140 SP = 140 × 80% = 112
PROFIT = 12 PROFIT % = 12%

3 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CHSL 2021

IF A NUMBER IS DECREASED BY 25% AND THEN INCREASED BY 25%, FIND NET CHANGE.

a) 5% DECREASE

b) 6.25% DECREASE

c) NO CHANGE

d) 6.25% INCREASE

EXPLANATION: NET % CHANGE = A − B − (AB/100)
= 25 − 25 − (625/100) = −6.25%

4 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CHSL 2019

AN ARTICLE IS SOLD AT 10% LOSS. IF SP = ₹450, FIND CP.

a) ₹520

b) ₹550

c) ₹480

d) ₹500

EXPLANATION: CP = SP × 100 / 90
= 450 × 100 / 90 = 500

5 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2022

A DISCOUNT OF 15% IS GIVEN ON AN ARTICLE. IF MARKED PRICE IS ₹800, FIND THE DISCOUNT AMOUNT.

a) ₹140

b) ₹100

c) ₹120

d) ₹160

EXPLANATION: DISCOUNT = 15% OF 800 = ₹120

6 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2021

MARKED PRICE OF AN ARTICLE IS ₹500. A DISCOUNT OF 10% IS GIVEN. FIND THE SELLING PRICE.

a) ₹470

b) ₹480

c) ₹450

d) ₹400

EXPLANATION: DISCOUNT = 10% OF 500 = 50
SP = 500 − 50 = ₹450

7 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2017

IF THE COST PRICE OF 20 ARTICLES EQUALS THE SELLING PRICE OF 16 ARTICLES, FIND PROFIT %.

a) 40%

b) 20%

c) 25%

d) 30%

EXPLANATION: CP OF 20 = SP OF 16
SO SP/CP = 20/16 = 1.25
PROFIT % = 25%

8 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2018

IF CP = ₹400 AND LOSS = 15%, FIND SP.

a) ₹340

b) ₹300

c) ₹360

d) ₹380

EXPLANATION: SP = 400 × 85% = ₹340

9 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CHSL 2022

A SHOPKEEPER MARKS AN ARTICLE AT ₹1000 AND ALLOWS 10% DISCOUNT. IF CP IS ₹800, FIND PROFIT %.

a) 12.5%

b) 15%

c) 10%

d) 20%

EXPLANATION: SP = 1000 × 90% = 900
PROFIT = 100 PROFIT % = (100/800) × 100 = 12.5%

10 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CHSL 2019

A MAN BUYS AN ARTICLE FOR ₹1200 AND SELLS IT AT ₹1500. FIND PROFIT %.

a) 20%

b) 30%

c) 35%

d) 25%

EXPLANATION: PROFIT = 300
PROFIT % = (300 / 1200) × 100 = 25%

11 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2020

IF THE SALARY OF A PERSON INCREASES FROM ₹8000 TO ₹9200, FIND THE PERCENTAGE INCREASE.

a) 18%

b) 15%

c) 12%

d) 20%

EXPLANATION: INCREASE = 9200 − 8000 = 1200
% INCREASE = (1200 / 8000) × 100 = 15%

12 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2019

IF SELLING PRICE IS DOUBLED AND COST PRICE REMAINS SAME, FIND PROFIT %.

a) 50%

b) 100%

c) 150%

d) 200%

EXPLANATION: PROFIT = SP − CP = CP
PROFIT % = 100%

13 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2022

A SHOPKEEPER GAINS 10% AFTER GIVING A DISCOUNT OF 20%. FIND MARKED PRICE IF COST PRICE IS ₹500.

a) ₹700

b) ₹750

c) ₹660

d) ₹687.5

EXPLANATION: SP = 500 × 110% = 550
MP = SP × 100 / 80 = 687.5

14 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CHSL 2021

A SHOPKEEPER ALLOWS A DISCOUNT OF 20% ON MARKED PRICE AND STILL GAINS 25%. FIND THE COST PRICE IF MP = ₹600.

a) ₹420

b) ₹400

c) ₹360

d) ₹384

EXPLANATION: SP = 600 × 80% = 480
CP = SP × 100 / 125 = 384

15 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2019

IF SP = ₹540 AFTER 10% DISCOUNT, FIND MP.

a) ₹580

b) ₹650

c) ₹620

d) ₹600

EXPLANATION: MP = SP × 100 / 90 = 600

16 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2018

IF A NUMBER IS INCREASED BY 30% AND THEN DECREASED BY 30%, WHAT IS THE NET CHANGE?

a) 9% INCREASE

b) NO CHANGE

c) 6% DECREASE

d) 9% DECREASE

EXPLANATION: NET % CHANGE = A − B − (AB/100)
= 30 − 30 − (900/100) = −9%

17 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CHSL 2020

A MAN BUYS AN ARTICLE FOR ₹800 AND SELLS IT AT A PROFIT OF 25%. FIND THE SELLING PRICE.

a) ₹1100

b) ₹1000

c) ₹950

d) ₹900

EXPLANATION: SP = CP × (100 + PROFIT%) / 100
= 800 × 125 / 100 = ₹1000

18 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC MTS 2021

SUCCESSIVE DISCOUNTS OF 15% AND 10% ARE EQUIVALENT TO:

a) 23.5%

b) 22%

c) 20%

d) 25%

EXPLANATION: NET DISCOUNT = 15 + 10 − (150/100) = 23.5%

19 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2020

SUCCESSIVE DISCOUNTS OF 10% AND 20% ARE EQUIVALENT TO A SINGLE DISCOUNT OF:

a) 32%

b) 28%

c) 25%

d) 30%

EXPLANATION: NET DISCOUNT = A + B − (AB/100)
= 10 + 20 − 2 = 28%

20 / 20

PERCENTAGE, PROFIT & LOSS AND DISCOUNT

SSC CGL 2020

IF A NUMBER IS INCREASED BY 50% AND THEN DECREASED BY 20%, FIND NET CHANGE.

a) 20% INCREASE

b) NO CHANGE

c) 5% DECREASE

d) 10% INCREASE

EXPLANATION: NET CHANGE = 50 − 20 − (1000/100) = 20%

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