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Delhi Police Exam Numerical Ability Study Material

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Arithmetic

Two trains start at the same time from A and B and proceed toward each other at the speed of 75 km/h and 50 km/h, respectively. When both meet at a point in between, one train found to be travelled 175 km more than the other. Find the distance between A and B.

a) 857
b) 875
c) 785
d) 758

✔Answer

✔Solution

Correct Answer Is : 875

Solution Is : According to the condition, = X/50=X+175/75
Where, x is the distance of the train which has speed is 50 km
75X=50X+50 x 175,25X =5870 ,x=350 ,Distance Between A and B is X +(X+175)=350+(350+175)=875

A candidate who gets 20% marks in an examination, fails by 30 marks. But if he gets 32 5 marks, he gets 42 marks more than the minimum pass marks. Find the pass percentage of marks:

a) 42%
b) 52%
c) 20%
d) 25%

✔Answer

✔Solution

Correct Answer Is : 25%

Solution Is : By condition 20% of x ? 30= 32% of x ? 42 - 20)%of x= 42 + 30 = 72
12% of x = 72
100% of x = 72 X 100 /12=600 ,Pass marks = 20% of 600 + 30 = (600 x 20)/100 + 30
= 120 +30= (150x100)/600
Per cent pass marks = 25%

Successive discounts of 20% and 10% are equivalent to single discount of

a) 30.00%
b) 15.00%
c) 28.00%
d) 25.00%

✔Answer

✔Solution

Correct Answer Is : 28.00%

Solution Is : Equivalent to a single discount of successive discount = (A+B)-(AxB)/100=28%

The average expenditure of a man for the first five months is Rs. 1200 and for the next seven months is Rs. 1300. If he saves Rs. 2900 in that year, his monthly average income is

a) Rs. 1400
b) Rs. 1500
c) Rs. 1600
d) Rs. 1700

✔Answer

✔Solution

Correct Answer Is : Rs. 1500

Solution Is : Average expenses of first 5 months = 1200
Average expenses of next 7 months = 1300
Saving of twetve months 2900
So, the total expenses of first 5 months = 6000
and total expenses of next 7 months = 9100
Sum of tweLve months= 15100
Saving of twelve months = 2900
Therefore, the total income of twelve months = 15100+ 2900
= 18000
Hence. the monthly income of a man = 12 = 1500

The vulgar fraction of 0.3939 is

a) (13/33)
b) (15/33)
c) (11/39)
d) (17/39)

✔Answer

✔Solution

Correct Answer Is : (13/33)

Solution Is : (p/q)=(3939-39)/9900=39/99=13/33

Merchant has 1000 kg sugar part of which he sells at 8°Io profit and the rest at 13% profit. He gains 14% on the whole. The quantity sold at 8°Io profit is

a) 400kg
b) 560kg
c) 600kg
d) 640kg

✔Answer

✔Solution

Correct Answer Is : 400kg

Solution Is : LetCP=C quantity of sugar on profit on 8%=x kg
1O8%of x X C + 118%of(1 000 -x) X C=114%of 1000 X C , 4000C=10 X C ,X=4000/10
Hence, required cuantity = 400kg

A man purchased an article for 1500 and sold it at 25% above the cost price. If he has to pay 75 as tax on it, his net profit percentage will be

a) 15
b) 20
c) 25
d) 30

✔Answer

✔Solution

Correct Answer Is : 20

Solution Is : The cost price of an article 1500
and the selling price of an articte= 125% of 1500= 1875
Amount of tax (pay)= 75
Then net prof (1875 ? 75 ? 1500) = 300
Per cent prof1= (300x100)/1500=20%

A number x is divisible by 7. When this number is divided by 8, 12 and 16, it leaves a remainder in each case. The least value of x is

a) 147
b) 148
c) 149
d) 150

✔Answer

✔Solution

Correct Answer Is : 147

Solution Is : LCM of 8. 12.16=48
Hence, the required number
= 48 x n + 3, dvisble by 7
= 48 x 3 + 3 = 147 divisible by 7 Let n =3

A man travels for 5 h 15 mm. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.

a) 240km
b) 1008 *6/7 km
c) 189km
d) 378km

✔Answer

✔Solution

Correct Answer Is : 240km

Solution Is : Let the total distance = d km According to the question. (d/2)/60+ (d/2)/45 =21/4,d=240 Km

If a man were to sell his hand-cart for Z 720, h would lose 25%. At what price must he sell it to gain 25%?

a) 768
b) 1200
c) 960
d) 1152

✔Answer

✔Solution

Correct Answer Is : 1200

Solution Is : Selling price of watch = 720 Loss = 25%
Let the cost price of watch = x
(100? 25)%O f X =720
125 x720 = 1200
125% of x = 125X720/75 =1200

Delhi Police Exam Numerical Ability Study Material

Arithmetic

Two trains start at the same time from A and B and proceed toward each other at the speed of 75 km/h and 50 km/h, respectively. When both meet at a point in between, one train found to be travelled 175 km more than the other. Find the distance between A and B.

Correct Answer Is : 875

Solution Is : According to the condition, = X/50=X+175/75 Where, x is the distance of the train which has speed is 50 km 75X=50X+50 x 175,25X =5870 ,x=350 ,Distance Between A and B is X +(X+175)=350+(350+175)=875

A candidate who gets 20% marks in an examination, fails by 30 marks. But if he gets 32 5 marks, he gets 42 marks more than the minimum pass marks. Find the pass percentage of marks:

Correct Answer Is : 25%

Solution Is : By condition 20% of x ? 30= 32% of x ? 42 - 20)%of x= 42 + 30 = 72 12% of x = 72 100% of x = 72 X 100 /12=600 ,Pass marks = 20% of 600 + 30 = (600 x 20)/100 + 30 = 120 +30= (150x100)/600 Per cent pass marks = 25%

Successive discounts of 20% and 10% are equivalent to single discount of

Correct Answer Is : 28.00%

Solution Is : Equivalent to a single discount of successive discount = (A+B)-(AxB)/100=28%

The average expenditure of a man for the first five months is Rs. 1200 and for the next seven months is Rs. 1300. If he saves Rs. 2900 in that year, his monthly average income is

Correct Answer Is : Rs. 1500

The vulgar fraction of 0.3939 is

Correct Answer Is : (13/33)

Solution Is : (p/q)=(3939-39)/9900=39/99=13/33

Merchant has 1000 kg sugar part of which he sells at 8°Io profit and the rest at 13% profit. He gains 14% on the whole. The quantity sold at 8°Io profit is

Correct Answer Is : 400kg

Solution Is : LetCP=C quantity of sugar on profit on 8%=x kg 1O8%of x X C + 118%of(1 000 -x) X C=114%of 1000 X C , 4000C=10 X C ,X=4000/10 Hence, required cuantity = 400kg

A man purchased an article for 1500 and sold it at 25% above the cost price. If he has to pay 75 as tax on it, his net profit percentage will be

Correct Answer Is : 20

Solution Is : The cost price of an article 1500 and the selling price of an articte= 125% of 1500= 1875 Amount of tax (pay)= 75 Then net prof (1875 ? 75 ? 1500) = 300 Per cent prof1= (300x100)/1500=20%

A number x is divisible by 7. When this number is divided by 8, 12 and 16, it leaves a remainder in each case. The least value of x is

Correct Answer Is : 147

Solution Is : LCM of 8. 12.16=48 Hence, the required number = 48 x n + 3, dvisble by 7 = 48 x 3 + 3 = 147 divisible by 7 Let n =3

A man travels for 5 h 15 mm. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.

Correct Answer Is : 240km

Solution Is : Let the total distance = d km According to the question. (d/2)/60+ (d/2)/45 =21/4,d=240 Km

If a man were to sell his hand-cart for Z 720, h would lose 25%. At what price must he sell it to gain 25%?

Correct Answer Is : 1200

Solution Is : Selling price of watch = 720 Loss = 25% Let the cost price of watch = x (100? 25)%O f X =720 125 x720 = 1200 125% of x = 125X720/75 =1200

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Solution Is : According to the condition, = X/50=X+175/75
Where, x is the distance of the train which has speed is 50 km
75X=50X+50 x 175,25X =5870 ,x=350 ,Distance Between A and B is X +(X+175)=350+(350+175)=875

Solution Is : By condition 20% of x ? 30= 32% of x ? 42 - 20)%of x= 42 + 30 = 72
12% of x = 72
100% of x = 72 X 100 /12=600 ,Pass marks = 20% of 600 + 30 = (600 x 20)/100 + 30
= 120 +30= (150x100)/600
Per cent pass marks = 25%

Solution Is : LetCP=C quantity of sugar on profit on 8%=x kg
1O8%of x X C + 118%of(1 000 -x) X C=114%of 1000 X C , 4000C=10 X C ,X=4000/10
Hence, required cuantity = 400kg

Solution Is : The cost price of an article 1500
and the selling price of an articte= 125% of 1500= 1875
Amount of tax (pay)= 75
Then net prof (1875 ? 75 ? 1500) = 300
Per cent prof1= (300x100)/1500=20%

Solution Is : LCM of 8. 12.16=48
Hence, the required number
= 48 x n + 3, dvisble by 7
= 48 x 3 + 3 = 147 divisible by 7 Let n =3

Solution Is : Selling price of watch = 720 Loss = 25%
Let the cost price of watch = x
(100? 25)%O f X =720
125 x720 = 1200
125% of x = 125X720/75 =1200