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SSC GD Constable 5 February 2025 · Shift 3 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 28 of 210

5 February 2025 · Shift 3

Elementary Mathematics

20 questions ~20 min readFree

20 questions

Find the value of

((63 \div 9) \times (\frac{14}{7} + \frac{19}{1} \times (7 - 6)))

A.150
B.148
C.159
D.147

Solution

BODMAS simplification Work from the innermost bracket outwards (BODMAS). Innermost bracket: (7-6)=1 Curly bracket:

\frac{14}{7} + \frac{19}{1} \times 1 = 2 + 19 = 21

Round bracket:

63 \div 9 = 7

Whole expression

= 7 \times 21 = 147

Hence the correct option is (d).

X is a greatest three-digit number exactly divisible by 12,20 , and 25 . The sum of the digits of X is:

Solution

LCM and digit sum A number divisible by 12, 20 and 25 must be a multiple of their LCM.

LCM

= 2^{2} \times 3 \times 5^{2} = 300

Greatest three-digit multiple of 300

= 300 \times 3 = 900

So X=900 and the sum of its digits =9+0+0=9 Hence the correct option is (b).

A retailer buys 12 pairs of shoes for ₹22,400. His overhead expenses are ₹1,600. If he earns ₹28,500 from the sale of these 12 pair of shoes, determine his profit percentage.

A.20%
B.17.5%
C.19.25%
D.18.75%

Solution

Profit percentage with overheads Overhead expenses are added to the cost price. Cost price = ₹22,400 + ₹1,600 = ₹24,000

Profit = ₹28,500 - ₹24,000 = ₹4,500 Profit percentage

= \frac{4500}{24000} \times 100

= 18.75 %

Hence the correct option is (d).

4 men and 3 women finish a job in 6 days, and 5 men and 7 women can do the same job in 4 days. How long will 8 men and 6 women take to do the same job ?

A.6 days
B.3 days
C.9 days
D.8 days

Solution

Time and work, men and women Let M be one man's one-day work and W one woman's one-day work.

The job is the same in both cases, so the two totals are equal. 6(4M+3W)=4(5M+7W) 24M+18W=20M+28W 4M=10W, so 1M=2.5W Total work

= 6 (4 \times 2.5 + 3) W = 6 \times 13 W = 78 W

One day's work of 8 men and 6 women

= 8 \times 2.5 W + 6 W = 26 W

Time

= \frac{78 W}{26 W} = 3

days Hence the correct option is (b).

The price of fuel decreases by 40%, 50% and 15% in three successive months, but increases by 40% in the fourth month. What is the percentage increase/decrease in the price of fuel in the fourth month as compared to its original price?

A.Decreases by 64.3%
B.Increases by 65.91%
C.Increases by 65.47%
D.Decreases by 69.09%

Solution

Successive percentage change Take the original price as ₹100 and apply the changes one after another.

After three falls:

100 \times \frac{60}{100} \times \frac{50}{100} \times \frac{85}{100} = 25.5

After the rise:

25.5 \times \frac{140}{100} = 35.7

Net change =35.7-100=-64.3 So the price falls by

64.3 %

. Hence the correct option is (a).

A car covers four successive 8 km distances at speeds of

12 km / hr, 18

km/hr, 24 km/hr and 36 km/hr, respectively. Its average speed (in

km / hr

) over this distance is:

A.19.2
B.20
C.18
D.22.5

Solution

Average speed, equal distances Average speed = total distance ÷ total time (never the average of the speeds). Total distance

= 4 \times 8 = 32

Total time

= \frac{8}{12} + \frac{8}{18} + \frac{8}{24} + \frac{8}{36} = \frac{6 + 4 + 3 + 2}{9}

hours

= \frac{15}{9} = \frac{5}{3}

hours Average speed

= 32 \div \frac{5}{3} = \frac{96}{5} = 19.2

km/hr Hence the correct option is (a).

The ratio of the walking speeds of

A

and

B

3 : 5

. If

B

takes 18 minutes to cover a certain distance, how much time will A take to cover the same distance ?

A.25 min
B.20 min
C.33 min
D.30 min

Solution

Speed and time, inverse ratio For the same distance, time is inversely proportional to speed. Speed ratio A:B=3:5, so time ratio A:B=5:3

\frac{t_{A}}{18} = \frac{5}{3}

t_{A} = \frac{5}{3} \times 18 = 30

minutes Hence the correct option is (d).

The third proportional to x and x + 100 is 405, find the value of x (where x > 100).

A.225
B.125
C.115
D.180

Solution

Third proportional If c is the third proportional to a and b, then a:b=b:c, so

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

. Here a=x, b=x+100 and c=405.

\frac{{(x + 100)}^{2}}{x} = 405

x^{2} + 200 x + 10000 = 405 x

x^{2} - 205 x + 10000 = 0

(x-125)(x-80)=0 x=125 or x=80; since x>100, x=125 Hence the correct option is (b).

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

\frac{3}{4}

of his normal speed, then he reaches City B 33 minutes late. Find the time (in minutes) that Chetan would have taken to travel from City A to City B if he drove at his normal speed,

A.97
B.100
C.99
D.93

Solution

Speed change and late arrival Speed becomes

\frac{3}{4}

of the normal speed, so the time becomes

\frac{4}{3}

of the normal time.

Let the normal time be T minutes.

\frac{4}{3} T - T = 33

\frac{T}{3} = 33

T=99 minutes Hence the correct option is (c).

If the radius of a hemisphere is 13

\sqrt{3} cm

, find its curved surface area.

A. $1014 π {cm}^{2}$
B. $1044 π {cm}^{2}$
C. $1024 π {cm}^{2}$
D. $1034 π {cm}^{2}$

Solution

Curved surface area of hemisphere Curved surface area of a hemisphere

= 2 π r^{2}

(the flat circular face is not counted).

r = 13 \sqrt{3}

r^{2} = {(13 \sqrt{3})}^{2} = 169 \times 3 = 507

CSA

= 2 π \times 507 = 1014 π

cm² Hence the correct option is (a).

If at same rate of interest, in 2 years, the simple interest is ₹40 and compound interest is ₹65, then what is the principal (in ₹) ?

A.11
B.16
C.20
D.9

Solution

Simple and compound interest Simple interest for 2 years is ₹40, so the interest of one year = ₹20.

CI - SI for 2 years = ₹65 - ₹40 = ₹25, which is one year's interest on the first year's ₹20.

20 \times \frac{R}{100} = 25

R=125

P = \frac{S I \times 100}{R \times T} = \frac{40 \times 100}{125 \times 2}

P = \frac{4000}{250} = 16

Hence the correct option is (b).

The average age of 39 students of a class is 39 years. If the age of the teacher is also included, the average age of the whole group becomes 40 years. The age (in years) of the teacher is:

A.81
B.77
C.79
D.83

Solution

Average age with one more member Total = average × number of members. Total age of 39 students

= 39 \times 39 = 1521

years Total age of 40 members

= 40 \times 40 = 1600

years

Teacher's age =1600-1521=79 years Hence the correct option is (c).

Evaluate

32 \div 8 \times 4 - 3 \times 4

Solution

BODMAS simplification Division and multiplication are done first, from left to right, and subtraction last.

32 \div 8 = 4

4 \times 4 = 16

3 \times 4 = 12

16-12=4 Hence the correct option is (c).

If the selling price of five Bluetooth devices is ₹14,430 after allowing a discount of

26 %

on the marked price, then the marked price of one device is:

A.₹3,000
B.₹3,900
C.₹3,365
D.₹2,886

Solution

Discount and marked price A 26% discount leaves the selling price at 74% of the marked price.

Marked price of 5 devices

= \frac{14430 \times 100}{74} = 19500

Marked price of one device

= \frac{19500}{5} = 3900

So one device is marked at ₹3,900. Hence the correct option is (b).

If the third proportional of

3 x^{2}

and 4 xy is 48, then the positive value of y is:

Solution

Third proportional in algebra The third proportional c of a and b satisfies

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

\frac{{(4 x y)}^{2}}{3 x^{2}} = 48

\frac{16 x^{2} y^{2}}{3 x^{2}} = 48

16 y^{2} = 144

y^{2} = 9

y=3 (positive value) Hence the correct option is (d).

The price of a computer is ₹22,200. What will be the price of the computer after reduction of 25% ?

A.₹16,550
B.₹16,750
C.₹16,650
D.₹16,450

Solution

Percentage reduction in price A fall of 25% leaves 75% of the price. New price

= \frac{75}{100} \times 22200

=16650 So the price becomes ₹16,650. Hence the correct option is (c).

Find the value

((42 \div 7) \times (\frac{42}{3} + \frac{19}{4} \times (8 - 4)))

A.180
B.209
C.198
D.202

Solution

BODMAS simplification Start with the innermost bracket. (8-4)=4, so

\frac{19}{4} \times 4 = 19

\frac{42}{3} = 14

Curly bracket =14+19=33

42 \div 7 = 6

Whole expression

= 6 \times 33 = 198

Hence the correct option is (c).

Ketan sold 151 chairs and had a gain equal to the selling price of 51 chairs. What is his profit percentage ?

A.56%
B. $46 %$
C. $61 %$
D.51%

Solution

Profit stated in selling price Let the selling price of one chair be s and its cost price be c. Gain on 151 chairs = selling price of 51 chairs, so 151s-151c=51s 100s=151c, i.e. the cost price of 151 chairs equals the selling price of 100 chairs.

Profit percentage

= \frac{151 s - 151 c}{151 c} \times 100 = \frac{51 s}{100 s} \times 100

= 51 %

Hence the correct option is (d).

Amit invests a sum of ₹5400 and Gopal invests a sum of ₹9400 at the same rate of simple interest per annum. If, at the end of 3 years, Gopal gets ₹840 more interest than Amit, then find the rate of interest per annum (in percentage).

Solution

Simple interest, difference method The rate and the time are the same for both, so the extra interest comes only from the extra principal. Extra principal = ₹9,400 - ₹5,400 = ₹4,000

\frac{4000 \times R \times 3}{100} = 840

120R=840 R=7 Hence the correct option is (b).

When difference between compound and simple interest for three years is ₹228 at 4% interest per annum, the principal is ₹

____

A.46875
B.48075
C.47295
D.46300

Solution

CI and SI difference, 3 years For 3 years, CI - SI

= P ({(1 + \frac{R}{100})}^{3} - 1 - \frac{3 R}{100}) = \frac{P R^{2} (300 + R)}{10^{6}}

Substituting R=4 and the difference 228:

228 = \frac{P \times 16 \times 304}{1000000}

228 = \frac{4864 P}{1000000}

P = \frac{228 \times 1000000}{4864}

P=46875 So the principal is ₹46,875. Hence the correct option is (a).

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