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SSC GD Constable 11 February 2025 · Shift 1 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 78 of 210

11 February 2025 · Shift 1

Elementary Mathematics

20 questions ~20 min readFree

20 questions

The price (per litre) of petrol increases by 85%. By what percent should its consumption be reduced such that the expenditure on it increases by 11% only ?

A.46%
B.60%
C.40%
D.53%

Solution

Percentage - price and consumption Expenditure = price × consumption, so fix price = 100 and consumption = 100 units. Original expenditure

= 100 \times 100 = 10000

. New price =100+85=185 and new expenditure

= 10000 \times \frac{111}{100} = 11100

. New consumption

= \frac{11100}{185} = 60

units.

Fall in consumption =100-60=40 units, i.e.

\frac{40}{100} \times 100 = 40

, so 40%. Hence, option (c) is correct.

The LCM of 30, 13, 180 and 234 is:

A.2409
B.2340
C.2379
D.2291

Solution

LCM by prime factorisation Break each number into primes.

30 = 2 \times 3 \times 5

, 13=13,

180 = 2^{2} \times 3^{2} \times 5

234 = 2 \times 3^{2} \times 13

The LCM takes the highest power of every prime that occurs. LCM

= 2^{2} \times 3^{2} \times 5 \times 13 = 4 \times 9 \times 5 \times 13 = 2340

. Hence, option (b) is correct.

Gopal invests a sum of ₹5400 and Akshay invests a sum of ₹10200 at the same rate of simple interest per annum. If, at the end of 4 years, Akshay gets ₹720 more interest than Gopal, then find the rate of interest per annum (in percentage).

A.3.75
B.5.75
C.2.75
D.1.75

Solution

Simple interest Formula: SI

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

. The rate and the time are the same for both, so the extra interest comes only from the extra principal. Extra principal =10200-5400=4800, i.e. ₹4,800.

\frac{4800 \times R \times 4}{100} = 720

192R=720

R = \frac{720}{192} = 3.75

So the required rate is 3.75% per annum. Hence, option (a) is correct.

When difference between compound and simple interest for three years is ₹93 at 10% interest per annum, the principal is ₹ ______.

A.3000
B.2425
C.3420
D.4200

Solution

CI − SI for three years Formula: for 3 years, CI - SI

= P ({(1 + \frac{R}{100})}^{3} - 1 - \frac{3 R}{100})

. With R=10:

{(\frac{11}{10})}^{3} = \frac{1331}{1000}

and

1 + \frac{3 \times 10}{100} = \frac{1300}{1000}

. So CI - SI

= P \times \frac{1331 - 1300}{1000} = \frac{31 P}{1000}

\frac{31 P}{1000} = 93

P = \frac{93 \times 1000}{31} = 3000

So the required principal is ₹3,000. Hence, option (a) is correct.

A man goes to Chennai from Hyderabad at a speed of 56 km/hr and returns to Hyderabad at a speed of 72 km/hr, through the same route. What is his average speed (in km/hr) of the entire journey?

A.60
B.62
C.61
D.63

Solution

Average speed both ways The distance each way is the same, so the plain average of the two speeds is wrong; use the harmonic mean. Formula: average speed

= \frac{2 x y}{x + y}

= \frac{2 \times 56 \times 72}{56 + 72}

= \frac{8064}{128}

=63 So the required average speed is 63 km/hr. Hence, option (d) is correct.

The LCM of 75, 45, 180 and 110 is:

A.9817
B.9856
C.9946
D.9900

Solution

LCM by prime factorisation Break each number into primes.

75 = 3 \times 5^{2}

45 = 3^{2} \times 5

180 = 2^{2} \times 3^{2} \times 5

110 = 2 \times 5 \times 11

LCM

= 2^{2} \times 3^{2} \times 5^{2} \times 11

= 4 \times 9 \times 25 \times 11 = 9900

Hence, option (d) is correct.

A person travels by a motorcycle to a place at an average speed of 60 km/h and returns at an average speed of 30 km/h. Find his average speed (in km/h) for the whole journey.

A.40
B.55
C.45
D.50

Solution

Average speed both ways Equal distances at two speeds, so use the harmonic mean, not

\frac{60 + 30}{2} = 45

. Formula: average speed

= \frac{2 x y}{x + y}

= \frac{2 \times 60 \times 30}{60 + 30}

= \frac{3600}{90}

=40 So the required average speed is 40 km/h. Hence, option (a) is correct.

A shopkeeper marks a table at

40 %

above the cost price and gives a discount of 25%. If he still makes a profit of ₹450, what is the cost price (in ₹) of the table ?

A.8,500
B.8,000
C.9,500
D.9,000

Solution

Mark-up, discount and profit Let the cost price be 100 units. Marked price =100+40=140 units. Selling price after 25% discount

= 140 \times \frac{75}{100} = 105

units. Profit =105-100=5 units, and this equals ₹450. 1 unit

= \frac{450}{5} = 90

Cost price

= 100 \times 90 = 9000

, i.e. ₹9,000. Hence, option (d) is correct.

The cost of painting a spherical vessel of radius 7 cm is ₹ 18,480 . What is the cost of painting per square centimetre ? (Use

π = \frac{22}{7}

)

A.₹33
B.₹32
C.₹30
D.₹31

Solution

Surface area of a sphere The paint covers the whole curved skin of the sphere.

Formula: surface area of a sphere

= 4 π r^{2}

= 4 \times \frac{22}{7} \times 7^{2}

= 4 \times 22 \times 7 = 616

​^{2}

Cost per cm

​^{2} = \frac{18480}{616} = 30

So the required cost is ₹30 per square centimetre. Hence, option (c) is correct.

LCM and HCF of two numbers are 144 and 8 respectively. If one of the numbers is 16 , then find the other number.

A.73
B.74
C.72
D.70

Solution

LCM × HCF relation Formula: LCM

\times

HCF = first number

\times

second number.

144 \times 8 = 16 \times x

1152=16x

x = \frac{1152}{16} = 72

Check: HCF of 16 and 72 is 8 and their LCM is 144, so the pair fits. Hence, option (c) is correct.

Determine the number that must be subtracted from both terms of the ratio 43 : 91 to make the ratio 3 : 7 .

Solution

Ratio with equal subtraction Let the number subtracted be x.

\frac{43 - x}{91 - x} = \frac{3}{7}

7(43-x)=3(91-x) 301-7x=273-3x 301-273=7x-3x 28=4x, so x=7. Check:

\frac{43 - 7}{91 - 7} = \frac{36}{84} = \frac{3}{7}

. Hence, option (c) is correct.

Ashok has ₹1612 with him. He divided it amongst his sons Raj and Varun and asked them to invest it at 8% rate of interest compounded annually. It was seen that Raj and Varun got same amount after 14 and 15 years respectively. How much (in ₹) did Ashok give to Raj ?

A.687
B.837
C.775
D.875

Solution

Compound interest - equal amounts Let Raj get ₹R and Varun get ₹V, so R+V=1612. Formula: Amount

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

; the two amounts are equal.

R {(\frac{27}{25})}^{14} = V {(\frac{27}{25})}^{15}

\frac{R}{V} = \frac{27}{25}

, so Raj's share is the larger one - his money grows for one year less.

27+25=52 units =1612, so 1 unit

= \frac{1612}{52} = 31

. Raj's share

= 27 \times 31 = 837

, i.e. ₹837. Hence, option (b) is correct.

Find the fourth proportional of 5√24, 3√10 and 3√15.

A.4.5
B.3
C.3.5
D.4

Solution

Fourth proportional with surds Formula: if a:b::c:d then

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

d = \frac{3 \sqrt{10} \times 3 \sqrt{15}}{5 \sqrt{24}}

= \frac{9 \sqrt{150}}{5 \sqrt{24}}

\sqrt{150} = 5 \sqrt{6}

and

\sqrt{24} = 2 \sqrt{6}

= \frac{9 \times 5 \sqrt{6}}{5 \times 2 \sqrt{6}} = \frac{45}{10}

=4.5 Hence, option (a) is correct.

The average age of 22 students of a class is 47 years. If the age of the teacher is also included, the average age of the whole group becomes 48 years. The age (in years) of the teacher is:

A.71
B.67
C.70
D.69

Solution

Average age Formula: total = average

\times

number of members. Total age of 22 students

= 22 \times 47 = 1034

years. Total age of 23 people

= 23 \times 48 = 1104

years. Teacher's age =1104-1034=70 years. Shortcut: the teacher raises 22 students' average by 1 year each, so his age

= 48 + 22 \times 1 = 70

. Hence, option (c) is correct.

Two trains are moving in opposite directions at speeds of 110 km/h and 140 km/h. The length of one train is 140 m. The time taken by them to cross each other is 5 seconds. The length (in m) of the other train, correct to 2 decimal places, is:

A.208.21
B.209.67
C.205.65
D.207.22

Solution

Trains crossing in opposite directions

Opposite directions, so relative speed =110+140=250 km/h. 250 km/h

= 250 \times \frac{5}{18} = \frac{625}{9}

m/s. In crossing, the two trains together cover the sum of their lengths. Sum of lengths

= \frac{625}{9} \times 5 = \frac{3125}{9} = 347.22

m. Other train =347.22-140=207.22 m. Hence, option (d) is correct.

The product of two numbers is 1,48,176 and their LCM is 3,528. What is their HCF ?

A.42
B.51
C.32
D.36

Solution

LCM × HCF relation Formula: product of two numbers = LCM

\times

HCF.

148176 = 3528 \times HCF

HCF

= \frac{148176}{3528}

=42 So the required HCF is 42. Hence, option (a) is correct.

The value of an ornament depreciates every year by 5%. If the present value of the ornament is ₹12,500, what will be its value after 2 years ?

A.₹11,315.50
B.₹11,250
C.₹11,325
D.₹11,281.25

Solution

Depreciation Depreciation is compound interest with a minus sign. Formula: value

= P {(1 - \frac{R}{100})}^{n}

= 12500 {(1 - \frac{5}{100})}^{2} = 12500 \times {(\frac{19}{20})}^{2}

= 12500 \times \frac{361}{400}

=11281.25 So the required value is ₹11,281.25. Hence, option (d) is correct.

By selling an article for ₹4,824, a man gains 20%. What will be his profit percentage (rounded off to the nearest integer) if he sells it for ₹6,000 ?

A.49%
B.53%
C.40%
D.32%

Solution

Profit percentage Formula: SP = CP

\times \frac{100 + profit}{100}

; the cost price never changes.

4824 = CP \times \frac{120}{100}

= \frac{4824 \times 100}{120} = 4020

, i.e. ₹4,020. New profit =6000-4020=1980, i.e. ₹1,980. Profit

% = \frac{1980}{4020} \times 100 = 49.25

Rounded to the nearest integer, 49%. Hence, option (a) is correct.

Kanchan sold 153 chairs and had a gain equal to the selling price of 78 chairs. What is his profit percentage?

A.104%
B.114%
C.109%
D.99%

Solution

Gain stated in selling price Let the selling price of one chair be s and its cost price be c. Profit = SP of 153 chairs - CP of 153 chairs = SP of 78 chairs. 153s-153c=78s 153s-78s=153c, so 75s=153c.

\frac{s}{c} = \frac{153}{75}

Profit

% = \frac{s - c}{c} \times 100 = (\frac{153}{75} - 1) \times 100 = \frac{78}{75} \times 100 = 104

So the required profit is 104%. Hence, option (a) is correct.

Ram, Ravi and Reena can do a piece of work in 16, 20 and 24 days, respectively. They began the work together but Ravi left the work 5 days before the completion of the work. In how much time did they finish the work together ?

A. $10 \frac{5}{34}$ days
B. $11 \frac{6}{37}$ days
C. $8 \frac{4}{37}$ days
D. $12 \frac{5}{48}$ days

Solution

Time and work - one leaves early Take total work = LCM of 16, 20 and 24 =240 units.

Let the work finish in T days. Ram and Reena work all T days; Ravi works (T-5) days. (15+10)T+12(T-5)=240 25T+12T-60=240 37T=300

T = \frac{300}{37} = 8 \frac{4}{37}

days. Hence, option (c) is correct.

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