For Parents For Educators & Experts For Institutions For Employers

SSC GD Constable 12 February 2025 · Shift 1 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 93 of 210

12 February 2025 · Shift 1

Elementary Mathematics

20 questions ~20 min readFree

20 questions

Find the ratio in which rice at ₹8.30 per kg be mixed with rice at ₹ 5.30 per kg to produce a mixture worth ₹6.30 per kg.

A. $2 : 1$
B.1 : 2
C. $1 : 3$
D. $2 : 3$

Solution

Alligation (mixtures) Apply the rule of alligation to the price of 1 kg: cheaper ₹5.30, dearer ₹8.30, mean ₹6.30.

Rule: (quantity of dearer) : (quantity of cheaper) = (mean - cheaper) : (dearer - mean). =(6.30-5.30):(8.30-6.30)=1.00:2.00 So the required ratio is 1:2 and the correct option is (b).

Zakir travels from City A to City B. If Zakir drives his car at

\frac{2}{3}

of his normal speed, then he reaches City B 45 minutes late. Find the time (in minutes) that Zakir would have taken to travel from City A to City B if he drove at his normal speed.

A.86
B.83
C.90
D.85

Solution

Speed, time and distance For a fixed distance, time is inversely proportional to speed. Speed becomes

\frac{2}{3}

of normal, i.e. speeds are in the ratio 3:2, so the times are in the ratio 2:3.

Difference of the two times =3-2=1 unit, and this delay is 45 minutes. Normal time =2 units

= 2 \times 45 = 90

minutes. So the correct option is (c).

P and Q can do a piece of work in 10 , and 6 days, respectively, while

R

can destroy the same work in 15 days. If P and Q work on

1^{st}

day, R works on

2^{nd}

and so on. They follow the same pattern until the work gets completed. How many days will it take for the work to be completed the first time?

A. $8 \frac{3}{4}$
B. $9 \frac{1}{2}$
C. $8 \frac{1}{2}$
D. $9 \frac{3}{4}$

Solution

Time and work (cycles) Take the total work as the LCM of 10, 6 and 15, i.e. 30 units. Then P does

\frac{30}{10} = 3

units a day, Q does

\frac{30}{6} = 5

units a day, and R destroys

\frac{30}{15} = 2

units a day. Day 1 (P and Q together) =3+5=8 units; day 2 (R alone) =-2 units, so one 2-day cycle gives 8-2=6 units.

After 8 days, i.e. 4 complete cycles, work done

= 4 \times 6 = 24

units, leaving 30-24=6 units. Day 9 is a P-and-Q day, and they do 8 units a day, so they need

\frac{6}{8} = \frac{3}{4}

of that day. Total time

= 8 + \frac{3}{4} = 8 \frac{3}{4}

days, so the correct option is (a).

Baban has two grandsons Rakesh and Shivam. 13 year old Rakesh gets some money from Baban's wealth and 14 year old Shivam gets rest of the money. But Rakesh and Shivam will get money only when they turn 22 years old. Till then the money is in a bank getting interest at rate 10% compounded annually. When both turn 22, they receive the same amount. How much had Baban given Shivam (in ₹) initially, if total money with Baban was ₹23100 ?

A.10750
B.12100
C.11000
D.12450

Solution

Compound interest (equal maturity) Rakesh is 13, so his share grows for 22-13=9 years; Shivam is 14, so his share grows for 22-14=8 years.

Let the shares be a and b. Equal amounts at 22 give

a {(\frac{11}{10})}^{9} = b {(\frac{11}{10})}^{8}

. Dividing both sides by

{(\frac{11}{10})}^{8}

a \times \frac{11}{10} = b

, so

\frac{a}{b} = \frac{10}{11}

. Hence a:b=10:11 and the total is 10+11=21 units = ₹23,100, so 1 unit = ₹1,100. Shivam's share

= 11 \times 1100 = 12100

, i.e. ₹12,100 - the correct option is (b).

Three persons step out together for a walk. Their steps measure 90 cm, 95 cm and 75 cm, respectively. What is the minimum distance (in m ) each should walk so that they can cover the distance in complete step?

A.85.05
B.87.06
C.87.6
D.85.5

Solution

LCM word problem Each man must cover a whole number of steps, so the distance must be a common multiple of 90, 95 and 75; the least such distance is their LCM.

90 = 2 \times 3^{2} \times 5

95 = 5 \times 19

75 = 3 \times 5^{2}

. Taking the highest power of each prime: LCM

= 2 \times 3^{2} \times 5^{2} \times 19 = 8550

cm. Converting, 8550 cm

= \frac{8550}{100} = 85.5

m. So the correct option is (d).

Evaluate

16 + 9 \div 3 - 2 \times 3

A.12
B.15
C.16
D.13

Solution

BODMAS simplification By BODMAS, division and multiplication are done before addition and subtraction.

16 + 9 \div 3 - 2 \times 3 = 16 + 3 - 6

=19-6=13 So the correct option is (d).

A scooterist is travelling at the speed of

42 km / h

. How much distance (in km ) will they travel in 90 minutes?

A.55
B.63
C.60
D.49

Solution

Distance from speed and time The speed is in km/h, so first change the time to hours: 90 minutes

= \frac{90}{60} = \frac{3}{2}

hours. Formula: Distance = Speed

\times

Time.

= 42 \times \frac{3}{2} = 63

So the distance is 63 km and the correct option is (b).

A man sold an article for ₹35,400 and made a profit of 18% in the process. What would have been his profit or loss percentage, had he sold it for ₹33,000?

A.Gain 15%
B.Loss 10%
C.Loss 15%
D.Gain 10%

Solution

Profit and loss percentage The cost price does not change, so find it first. A profit of 18% means SP

= \frac{118}{100} \times

CP, so CP

= 35400 \times \frac{100}{118} = 30000

, i.e. ₹30,000. With the new selling price ₹33,000, gain =33000-30000=3000. Gain percentage

= \frac{3000}{30000} \times 100 = 10

. So he would gain 10% and the correct option is (d).

The value of

68 - 6 \times [4 + 4 \times {12 - 4

(6 - 5) \times 3} \div 15

] is :

A.54
B.50
C.47
D.44

Solution

BODMAS with brackets Simplify from the innermost bracket outwards. Inside the braces:

12 - 4 (6 - 5) \times 3 = 12 - 4 \times 1 \times 3 = 12 - 12 = 0

. Now the square bracket:

4 + 4 \times 0 \div 15 = 4 + 0 = 4

. Finally

68 - 6 \times 4 = 68 - 24 = 44

. So the correct option is (d).

The average weight of Amit, Gopal and Akshay is 45 kg . If the average weight of Amit and Gopal be 43 kg and that of Gopal and Akshay be 47 kg, then the weight of Gopal (in kg) is:

A.65
B.55
C.60
D.45

Solution

Averages of groups Sum = average

\times

number of persons. Amit + Gopal + Akshay

= 3 \times 45 = 135

kg. Amit + Gopal

= 2 \times 43 = 86

kg. Gopal + Akshay

= 2 \times 47 = 94

kg. Adding the last two: Amit + 2 Gopal + Akshay =86+94=180 kg. Subtracting the first sum removes Amit and Akshay: Gopal =180-135=45 kg. So the correct option is (d).

When

8, 16, 18, 20

and 25 divide the least number

x

, the remainder in each case is 3 , but

x

is divisible by 7 . What is the value of x ?

A.7203
B.7302
C.7320
D.7023

Solution

LCM with common remainder A number that leaves remainder 3 with each divisor is (a common multiple of the divisors) +3.

8 = 2^{3}

16 = 2^{4}

18 = 2 \times 3^{2}

20 = 2^{2} \times 5

25 = 5^{2}

, so LCM

= 2^{4} \times 3^{2} \times 5^{2} = 3600

. Hence x=3600k+3 for some positive integer k. k=1 gives x=3603, and

3603 = 7 \times 514 + 5

, so it is not divisible by 7. k=2 gives x=7203, and

7203 = 7 \times 1029

exactly. So the least such number is x=7203 and the correct option is (a).

The expenditure of Sudha is

175 %

more than her savings. If her expenditure decreases by 12% and savings increase by 40.5%, then by what percent does her income increase?

A.4%
B.5%
C.2%
D. $7 %$

Solution

Percentage change in income Let the savings be 100. Expenditure is 175% more than savings, so expenditure =100+175=275. Income = expenditure + savings =275+100=375. New expenditure (12% less)

= 275 \times \frac{88}{100} = 242

. New savings (40.5% more)

= 100 \times \frac{140.5}{100} = 140.5

. New income =242+140.5=382.5, so the rise is 382.5-375=7.5. Increase percentage

= \frac{7.5}{375} \times 100 = 2

. So the income increases by 2% and the correct option is (c).

The simple interest on a principal amount (in ₹) is ₹ 531 for a period of 5 years at the rate of 9% per annum. The principal amount (in ₹) is:

A.1171
B.1186
C.1179
D.1180

Solution

Simple interest principal Formula: SI

= \frac{P \times R \times T}{100}

531 = \frac{P \times 9 \times 5}{100}

P = \frac{531 \times 100}{45}

P=1180 So the principal is ₹1,180 and the correct option is (d).

The height of a right circular cone is 10 cm and its base radius is 24 cm . What is the curved surface area (in

{cm}^{2}

) of the cone?

A. $640 π$
B. $620 π$
C. $624 π$
D. $580 π$

Solution

Curved surface area of a cone The curved surface area needs the slant height, not the vertical height.

Slant height:

l = \sqrt{h^{2} + r^{2}} = \sqrt{10^{2} + 24^{2}} = \sqrt{100 + 576} = \sqrt{676} = 26

cm. Formula: CSA

= π r l

= π \times 24 \times 26 = 624 π

So the curved surface area is

624 π

​^{2}

and the correct option is (c).

The population of a town increases by 8% annually. If the present population is 45,000 , what will be its population after 2 years?

A.52,488
B.52,200
C.52,316
D.52,532

Solution

Population growth (compound) Annual growth compounds, so use

P_{n} = P {(1 + \frac{r}{100})}^{n}

= 45000 \times {(\frac{108}{100})}^{2} = 45000 \times \frac{27}{25} \times \frac{27}{25}

= 48600 \times \frac{27}{25}

=52488 So the population after 2 years is 52,488 and the correct option is (a).

The third proportional of 20 and 25 is:

A.31.75
B.32.75
C.31.25
D.32.25

Solution

Third proportional If c is the third proportional to a and b, then a:b=b:c, i.e.

c = \frac{b^{2}}{a}

. Here a=20 and b=25.

c = \frac{25^{2}}{20} = \frac{625}{20}

=31.25 So the correct option is (c).

The present age ratio of the mother to the daughter is

7 : 2

. After 12 years, their ages will be in the ratio of 9 : 4. What is the current age of the mother?

A.48 years
B.52 years
C.42 years
D.38 years

Solution

Ages and ratios Let the present ages be 7x and 2x years. After 12 years:

\frac{7 x + 12}{2 x + 12} = \frac{9}{4}

. Cross-multiplying: 4(7x+12)=9(2x+12) 28x+48=18x+108 10x=60, so x=6. Mother's present age

= 7 \times 6 = 42

years, so the correct option is (c).

The marked price of a table is ₹8,980. If the shopkeeper offers a discount of 19%, what is its selling price (in ₹, to the nearest rupee) ?

A.₹7,427
B.₹7,472
C.₹7,247
D.₹7,274

Solution

Discount and selling price A discount of 19% is given on the marked price, so SP = MP

\times \frac{100 - 19}{100}

= 8980 \times \frac{81}{100}

=7273.8 Rounded to the nearest rupee, SP = ₹7,274, so the correct option is (d).

400 guavas were bought at ₹1240 per hundred and were sold at a profit of ₹940. Find the selling price (in ₹) per dozen of guavas.

A.167
B.192
C.177
D.187

Solution

Profit and unit selling price CP of 100 guavas is ₹1,240, and 400 guavas is 4 such hundreds. CP of 400 guavas

= 4 \times 1240 = 4960

, i.e. ₹4,960. SP of 400 guavas = CP + profit =4960+940=5900, i.e. ₹5,900. SP of one guava

= \frac{5900}{400} = 14.75

SP per dozen

= 14.75 \times 12 = 177

So the correct option is (c).

Baban has two grandsons Rakesh and Shivam. 13 year old Rakesh gets some money from Baban's wealth and 14 year old Shivam gets rest of the money. But Rakesh and Shivam will get money only when they turn 22 years old. Till then the money is in a bank getting interest at rate 5% compounded annually. When both turn 22, they receive the same amount. How much had Baban given Shivam (in ₹) initially, if total money with Baban was ₹24600?

A.12600
B.11750
C.12000
D.12950

Solution

Compound interest (equal maturity) Rakesh's share is invested for 22-13=9 years and Shivam's for 22-14=8 years at 5% per annum.

Let the shares be a and b. Equal amounts at 22 give

a {(\frac{21}{20})}^{9} = b {(\frac{21}{20})}^{8}

. Dividing both sides by

{(\frac{21}{20})}^{8}

a \times \frac{21}{20} = b

, so

\frac{a}{b} = \frac{20}{21}

. Hence a:b=20:21 and the total is 20+21=41 units = ₹24,600, so 1 unit = ₹600. Shivam's share

= 21 \times 600 = 12600

, i.e. ₹12,600 - the correct option is (a).

Finished reading? Now test yourself.

The same 20 questions, but with a timer and the answers hidden — 15 minutes.

Take the test

Full contents list