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SSC GD Constable 17 February 2025 · Shift 2 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 128 of 210

17 February 2025 · Shift 2

Elementary Mathematics

20 questions ~20 min readFree

20 questions

If 4% of x=108, then x is equal to:

A.2700
B.5500
C.5400
D.2800

Solution

Percentage of a number Let the required number be x; the question says 4% of it is 108.

\frac{4}{100} \times x = 108

x = 108 \times \frac{100}{4}

x = 27 \times 100 = 2700

Hence x = 2700, so option (a) is correct.

A soap company produces two kinds of soaps having weight 500 gm and 650 gm , respectively. If 200 soaps of the first kind and 300 soaps of second the kind were produced, then what is the weighted average weight (in gm ) of the soaps?

A.620
B.590
C.610
D.600

Solution

Weighted average Formula: weighted average = (total weight)

\div

(total number of items).

Total weight

= 200 \times 500 + 300 \times 650 = 100000 + 195000 = 295000

gm Total number of soaps =200+300=500 Weighted average

= \frac{295000}{500} = 590

gm Hence the weighted average weight is 590 gm, so option (b) is correct.

LCM and HCF of two numbers are 140 and 5 respectively. If one of the numbers is 35 , then find the other number.

A.23
B.19
C.22
D.20

Solution

LCM and HCF relation Formula: LCM

\times

HCF = product of the two numbers. Let the other number be N.

140 \times 5 = 35 \times N

700 = 35N

N = \frac{700}{35} = 20

Check: HCF of 35 and 20 is 5 and their LCM is 140, so option (d) is correct.

Find the value of [(29 × 13) × {3 ÷ 3 × (18 − 14)/4}]

A.380
B.377
C.364
D.391

Solution

BODMAS simplification BODMAS: clear the innermost bracket first, then the curly bracket, then multiply. Innermost bracket: 18-14=4, so

\frac{18 - 14}{4} = \frac{4}{4} = 1

. Curly bracket:

3 \div 3 \times 1 = 1 \times 1 = 1

. Round bracket:

29 \times 13 = 377

. Now

377 \times 1 = 377

. Hence the value is 377, so option (b) is correct.

Evaluate:

16 + 12 \div 3 - 3 \times 3

A.14
B.10
C.11
D.13

Solution

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

12 \div 3 = 4

and

3 \times 3 = 9

16+4-9 = 20-9 = 11 Hence the value is 11, so option (c) is correct.

The value of 69 − 4 × [4 + 3 × {20 − 5(7 − 5) × 2} ÷ 29] is:

A.54
B.45
C.46
D.53

Solution

BODMAS simplification Work outward: round bracket, then curly bracket, then square bracket. Round bracket: 7-5=2, so

5 (7 - 5) \times 2 = 5 \times 2 \times 2 = 20

. Curly bracket: 20-20 = 0. Next,

3 \times 0 \div 29 = 0

. Square bracket: 4+0 = 4.

69 - 4 \times 4 = 69 - 16 = 53

Hence the value is 53, so option (d) is correct.

Shivam has two grandsons Rupesh and Mandar 12 year old Rupesh gets some money from Shivam's wealth and 13 year old Mandar gets rest of the money. But Rupesh and Mandar will get money only when they turn 23 years old. Till then the money is in a bank getting interest at rate 8% compounded annually. When both turn 23 , they receive the same amount. How much had Shivam given Mandar (in ₹) initially, if total money with Shivam was ₹24700 ?

A.11875
B.13175
C.11625
D.12825

Solution

Compound interest, equal amounts Rupesh is 12, so his money grows for 23-12=11 years; Mandar is 13, so his grows for 23-13=10 years. Formula: Amount

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

Both amounts are equal, so

P_{R} {(\frac{27}{25})}^{11} = P_{M} {(\frac{27}{25})}^{10}

(since

1 + \frac{8}{100} = \frac{27}{25}

). Cancelling

{(\frac{27}{25})}^{10}

gives

P_{R} \times \frac{27}{25} = P_{M}

, i.e.

\frac{P_{R}}{P_{M}} = \frac{25}{27}

P_{R} : P_{M} = 25 : 27

and 25+27=52 units = ₹24,700. 1 unit

= \frac{24700}{52} = 475

Mandar's share

= 27 \times 475 = 12825

Hence Shivam gave Mandar ₹12,825, so option (d) is correct.

A and B can do a piece of work in 18 days, B and C can do the same work in 32 days and

A

and

C

can do the same in 52 days. The time taken by A, B and C together to do the work is:

A. $18 \frac{342}{397}$ days
B. $17 \frac{341}{397}$ days
C. $15 \frac{339}{397}$ days
D. $16 \frac{340}{397}$ days

Solution

Time and work, pairs Take one day’s work:

A + B = \frac{1}{18}

B + C = \frac{1}{32}

A + C = \frac{1}{52}

. Adding the three:

2 (A + B + C) = \frac{1}{18} + \frac{1}{32} + \frac{1}{52}

LCM of 18, 32 and 52 is 3744, so the sum

= \frac{208 + 117 + 72}{3744} = \frac{397}{3744}

A + B + C = \frac{397}{7488}

of the work per day Time

= \frac{7488}{397}

days

397 \times 18 = 7146

and 7488-7146=342, so the time

= 18 \frac{342}{397}

days Hence option (a) is correct.

Nirmal invests a sum of ₹5400 and Rakesh invests a sum of ₹9400 at the same rate of simple interest per annum. If, at the end of 5 years, Rakesh gets ₹360 more interest than Nirmal, then find the rate of interest per annum (in percentage).

A.3.2
B.3.8
C.1.8
D.2.2

Solution

Simple interest, difference Formula:

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

. Rate and time are the same for both, so the difference of interests comes only from the difference of principals. Difference of principals =9400-5400=4000

\frac{4000 \times R \times 5}{100} = 360

200R=360

R = \frac{360}{200} = 1.8

Hence the rate is 1.8% per annum, so option (c) is correct.

Find the area of a triangle, whose sides are

6 cm, 0.08 m

and 4 cm .

A. $5 \sqrt{15} {cm}^{2}$
B. $4 \sqrt{15} {cm}^{2}$
C. $3 \sqrt{15} {cm}^{2}$
D. $6 \sqrt{15} {cm}^{2}$

Solution

Heron's formula First bring the sides to one unit: 0.08 m

= 0.08 \times 100 = 8

cm. So the sides are 6 cm, 8 cm and 4 cm.

Semiperimeter

s = \frac{6 + 8 + 4}{2} = \frac{18}{2} = 9

cm Formula (Heron's): Area

= \sqrt{s (s - a) (s - b) (s - c)}

Area

= \sqrt{9 (9 - 6) (9 - 8) (9 - 4)} = \sqrt{9 \times 3 \times 1 \times 5}

Area

= \sqrt{135} = \sqrt{9 \times 15} = 3 \sqrt{15}

​^{2}

Hence option (c) is correct.

Which of the following numbers will come in the place of x in the given equation? 0.3 : 4.8 : : 0.2 : x

A.3.8
B.2.1
C.3.2
D.4.2

Solution

Proportion In a proportion a:b::c:d, the ratios are equal, so

\frac{a}{b} = \frac{c}{d}

\frac{0.3}{4.8} = \frac{0.2}{x}

Cross-multiplying:

0.3 \times x = 4.8 \times 0.2

0.3x = 0.96

x = \frac{0.96}{0.3} = 3.2

Shortcut:

4.8 = 16 \times 0.3

, so

x = 16 \times 0.2 = 3.2

; option (c) is correct.

Which scheme provides the highest discount among the options given?

A.Two successive discounts of 8% each
B.Two successive discounts of 10% each
C.Two successive discounts of 5% each
D.Two successive discounts of 8% and 11%

Solution

Successive discounts Formula: two successive discounts of

a %

and

b %

equal one discount of

(a + b - \frac{a b}{100}) %

. (a)

8 + 8 - \frac{8 \times 8}{100} = 16 - 0.64 = 15.36 %

(b)

10 + 10 - \frac{10 \times 10}{100} = 20 - 1 = 19 %

(c)

5 + 5 - \frac{5 \times 5}{100} = 10 - 0.25 = 9.75 %

(d)

8 + 11 - \frac{8 \times 11}{100} = 19 - 0.88 = 18.12 %

The largest single equivalent discount is

19 %

, so option (b) gives the highest discount.

A train 268 m long is moving at a speed of

58 km / h

. How much time will it take to cross a man coming from the opposite direction at a speed of 14 km/h?

A.12.8 seconds
B.17.2 seconds
C.15.5 seconds
D.13.4 seconds

Solution

Train crossing a man They move in opposite directions, so the speeds add up. Relative speed =58+14=72 km/h 72 km/h

= 72 \times \frac{5}{18} = 20

m/s A man is treated as a point, so the train covers only its own length, 268 m. Time

= \frac{distance}{relative speed} = \frac{268}{20} = 13.4

seconds Hence option (d) is correct.

The price (per litre) of petrol increases by 50%. By what percent should its consumption be reduced such that the expenditure on it increases by 41% only ?

A.6%
B.8%
C.94%
D.93%

Solution

Price, consumption and expenditure Expenditure = price

\times

consumption. Let the original price and consumption both be 100, so the expenditure

= 100 \times 100 = 10000

. New price

= 100 + 50 %

of 100=150 New expenditure

= 10000 + 41 %

of 10000=14100 New consumption

= \frac{14100}{150} = 94

Fall in consumption =100-94=6 Percentage reduction

= \frac{6}{100} \times 100 = 6 %

Hence option (a) is correct.

A man sold an article for ₹219, there by gaining 9.5%. The cost of the article (in ₹) was:

A.239
B.198
C.241
D.200

Solution

Cost price from gain percent Formula:

S P = C P (1 + \frac{gain %}{100})

219 = C P \times (1 + \frac{9.5}{100})

219 = C P \times \frac{109.5}{100} = C P \times \frac{219}{200}

C P = 219 \times \frac{200}{219} = 200

Hence the cost price is ₹200, so option (d) is correct.

Pravin has ₹1224 with him. He divided it amongst his sons Rishi and Shan and asked them to invest it at 4% rate of interest compounded annually. It was seen that Rishi and Shan got same amount after 15 and 16 years respectively. How much (in ₹) did Pravin give to Shan ?

A.600
B.724
C.450
D.624

Solution

Compound interest, equal amounts Rishi's money grows for 15 years and Shan's for 16 years at 4% per annum, and

1 + \frac{4}{100} = \frac{26}{25}

. Equal amounts:

P_{R} {(\frac{26}{25})}^{15} = P_{S} {(\frac{26}{25})}^{16}

Cancelling

{(\frac{26}{25})}^{15}

gives

P_{R} = P_{S} \times \frac{26}{25}

, so

\frac{P_{R}}{P_{S}} = \frac{26}{25}

. So

P_{R} : P_{S} = 26 : 25

and 26+25=51 units = ₹1,224. 1 unit

= \frac{1224}{51} = 24

Shan's share

= 25 \times 24 = 600

Hence Pravin gave Shan ₹600, so option (a) is correct.

The average of 21 numbers is 74 . The average of the first 11 numbers is 72.8 and that of the last 11 numbers is 77.2 . If the 10th number is five less than the 11th number and if 10th and 11th numbers are excluded, then what is the average of the remaining numbers ? (correct to nearest integer).

A.71
B.70
C.73
D.72

Solution

Average, overlapping groups Sum of all 21 numbers

= 21 \times 74 = 1554

Sum of the first 11 numbers

= 11 \times 72.8 = 800.8

Sum of the last 11 numbers

= 11 \times 77.2 = 849.2

The two blocks together cover 22 places, so the 11th number is counted twice. 11th number =(800.8+849.2)-1554=1650-1554=96 10th number =96-5=91 Sum of the remaining 19 numbers =1554-96-91=1367 Average

= \frac{1367}{19} = 71.9 \dots \approx 72

Hence option (d) is correct.

Baban sold 150 chairs and had a gain equal to the selling price of 90 chairs. What is his profit percentage ?

A.145%
B.150%
C.155%
D.160%

Solution

Profit percentage Let the cost price of one chair be c and the selling price of one chair be s. Profit on 150 chairs =150s-150c, and this equals the selling price of 90 chairs =90s. 150s-150c=90s 60s=150c, so

\frac{s}{c} = \frac{150}{60} = \frac{5}{2}

Take s=5 and c=2; profit per chair =5-2=3. Profit

% = \frac{3}{2} \times 100 = 150 %

Hence option (b) is correct.

x

varies indirectly to

y^{2}

and if

x =

78,

y = 11

, then find the value of x when y

= 2.2

A.1950
B.1995
C.1880
D.1725

Solution

Inverse variation Inverse variation with

y^{2}

means

x \propto \frac{1}{y^{2}}

, i.e.

x y^{2} = k

(a constant). From the given values:

k = 78 \times 11^{2} = 78 \times 121 = 9438

When y=2.2:

y^{2} = 2.2 \times 2.2 = 4.84

x = \frac{9438}{4.84}

x=1950 Hence option (a) is correct.

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

\frac{3}{7}

of his normal speed, then he reaches City B 28 minutes late. Find the time (in minutes) that Bhanu would have taken to travel from City A to City B if he drove at his normal speed.

A.18
B.11
C.21
D.19

Solution

Speed, time and delay For a fixed distance, time varies inversely with speed. Speed becomes

\frac{3}{7}

of normal, so time becomes

\frac{7}{3}

of the normal time T. Extra time

= \frac{7}{3} T - T = \frac{4}{3} T

\frac{4}{3} T = 28

T = 28 \times \frac{3}{4} = 21

minutes Hence option (c) is correct.

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