Hints and Solutions to

Mathematics Speed Tests

MT 35 – 50

MT 35

Solutions

1. (5) None of these

→ 27 × ? = 4779 = 177

2. (1) 8.225

3. (4) 67

By approximation method,

100 × ? % of 120 = 8104

? % of 120 = 8104

? % = 67

4. (4) 4

5. (1) 22 11/30

6. (4) 100

By approximation;

= 100

7. (3) 31200

By approximation;

8. (4) 1020

465 + 554 1020

9. (1) 380

151.1 + (151.1 + 76)

10. (2) 700

11. (5) 36

By approximation;

50 – 35 + 21 (nearest roots)

12. (3) 73

MT 36

Solutions

1. 50% + 0% = 95%

95% + 12% = 118.4%

2. 15% + 5% +

20.75%

3.

x = 220

4. (75% + 45%) – 100% = 20%

20% plays both; so No. of student = 600 × 20% = 120

5. (

) - 132000 = 15840 votes

6. 110 – (3 × 4 + 8 × 6) =

7. Reduced % = 100 – 92 = 8%

As, if 125 is the price, he spends 115 i.e, 125 → 115

if 100 is price he spends ? = 92 100 → ? = 92

8.

Rs. 4/kg

80 → Rs. 4

100 → ? = Rs. 5/kg (Original Price)

9. +30% - 30% = - 9%

After increasing by 10%

- 9% + 10% = 0.1% more from January.

10.

x = 450 × 40% = 180

MT 37

Solutions

1. – 50% - 30% = 65%

- 65% - 12% = 69.2% discount

2. 15 12

12 15

144 225

3.

profit

4. 324 – 24 = 300 is paid by C to B;

gain

150 → 300

100 → ? = 200

5. 15% - 10% + 1.5% = 23.5% discount

76.5 → 306

100 → ? = 400

6.

7. 80 → 140

100 → ? = 175 (M.P.)

125 → 140

100 → ? = 112 (C.P.)

8. a. C.P. 36, Profit 17 = 47.22% profit

b. C.P. 50, Profit 24 = 48% profit

c. C.P. 40, Profit 19 = 47.5% profit

d. C.P. 60, Profit 29 = 48.33% profit

So, Best Transaction is (D)

9. 80 → 9

100 → ? = 135

10. 105 – 95 = 10

10 → 15

100 → ? = 150

11.

12.

MT 38

Solutions

1. 900 + 60 + 132 = 1092 (interest)

A = 1500 + 1092 = 2592

OR, 1500

= 2592

2.

(Interest)

A = 7200 + 4860 = 12060

3. 140 → 2520

100 → ? = 1800

4. Expected interest = 100;

years

5.

years

6. 850 = P

; P = 85000

7. 3x = 9; 3

x =3

2

So, x = 2 → 2 × 3 years = 6 years

8. Per year Int. =

; Sum (P) = 250

9. 40 = ? % of 800 ; r = 5% [ interest on interest for a year = 840 - 800 = 40]

10. 2.5% × 4 years = 10% more

10% → 1400

100% → ? = 14000

MT 39

Solutions

1. C – D = 4 – 3 = 1

If, 1 part = 1000

2 part = 2000 (B’s share)

2. Third No. = 100; then,

3. x =

4. 6 : 8 : 9

First part =

5.

6.

7. Other than option (B) has multiples for 12 (A dozen).

8.

9. P : Q : R = (5 × 3) : (7 × 3) : (5 × 9) = 5 : 7 : 15

10. 8A = 12B = 6C;

So, A : B : C = 3 : 2 : 4 ; A’s share =

11.

So,

12.

Total property = 9000 × 20 = 180000

OR,

50% + 25% + 20% = 95%

Remaining 5% = 9000

100% = ? = 180000

MT 40

Solutions

1. (27 × 5) – (25 × 4) = 35

2. 35.4 + (0.4) x 24 = 45 kgs

3. 45 + (0.2) × 60 = 57 kgs

4.

i.e,

=

5. 32 + (7x – 2) + (7x + 2) = 32

6. Total age 3 years ago of all = 27 x 3 = 81

Total present Age of all = 81 + 3x3 = 90 years

Total age of Wife and Child now = 40 + 2x5= 50 years

So, Husband’s Age = 90 – 50 = 40 years

7.

Students

8.

litres consumed in 3 years.

So, Average cost =

9. Boys =

Girls = 6

Average age of boys = twice the No. of Girls = 2 × 6 = 12 years

So, total age = 12 × 30 = 360 years

10. 120 (20 × 9) = 300 gms

11. 60 + [5 x (+ 2)] + [5 x (+ 4)] = 90

MT 41

Solutions

1.

i.e., In equal Quality. So, he sells 25 pens profit and 25 pens at loss.

2.

i.e. In ratio of 1 : 3, tea should be mixed.

3.

litres

4. In ratio of 15 : 100

i.e., 15% part is adulterated.

5.

So, x = Rs. 30 / kg

6.

So, sugar is saled in 2 : 3 ratio. Quantity at 18% profit =

Kgs

7. C.P. per kg of sugar mixture =

If 110 → 9.24

100 → ? = 8.40 / kg

So, sugar is mixed in 7 : 3 ratio. Quantity of sugar @ 9 per kg = 63 kgs.

8. M W

3 : 2

50 litres → 30 20

For equal quantity in mixture, 10 litres of water should be added.

9. M W

So, milk and water should be in ratio of 2 : 1.

10. Each student’s average share =

B G

Boys

and girls are in ratio of 3 : 2. So, No. of girls =

.

MT 42

Solutions

1. T = 2000 × 9 = 18000

P = 5000 × 7 = 35000

T : P = 18 : 35

2. (5 × 14) : (7 × 8) : (8 ×7) = 5 : 4 : 4

3.

4. B’s share as an active partner = 8000 × 10% = 800

Remaining Part’s share of B =

B’s total profit = 2700 + 800 = 3500

5. Investment Ratio of A : B : C =

[(20000 x5) + (15000 x7) ] : [(20000 x5) + (16000 x7)] : [(2000 x5) + (26000 x7)] =

205:212:282

Share of A =

So, as B = 21200

and C = 28200

6. B claims

so A get =

So,

x = 12800

7. A : B : C = 60 : 40 : 24 = 15 : 10 : 6

So, C’s share =

8.

Total profit after charity = 1425

If 95 → 1425

100 → ? = 1500

9. A : G = (10 × 5) : (8 × 10) = 15 : 16

Gama’s share =

10. Investment Ratio = E : M : D = (5 x10) : (3 x 10) : (4 x 8) = 25 : 15 : 16

So, Mina’s share (Profit) =

x 168000 = 45000.

MT 43

Solutions

1. Second No. =

2. L.C.M of 12, 18, 21, 30 = 1260

Given, number that is doubled = 1260

So least number will be =

= 630

3. L.C.M. = 3 × 4 × 4 = 48

4. Suppose HCF =

Then, 2 x 3 = 48 ; so, = 8

Sum of Numbers = 2x8 + 3x8 = 40.

5. L.C.M. of 126, 84 and 78 = 3276 sec

6. 1657 – 6 = 1651

2037 – 5 = 2032

H.C.F = 127

7. 2 + 2 + 3 + * + 4 + 3 + 1 = 15 + * ;

so * = 3 as 15+3 = 18 is divisible by 3.

8.

Remainder = 21 ; So least No. = 2

9. Smaller No. = , so Larger No. = + 1365

+ 1365 = 6 +15

= 270

10. 2 (1 + 2 + 3 + 4 + ….. 41)

11. 80 = 10 x 8

According to Divisibility rule of 10, at the place of y, there will be 0.

And according to divisibility rule of 8, last three digit should be divisible by 8.

i.e, 3 + + 0, should be divisible by 8. So, in place of there should be 2.

So, + y = 2 + 0 = 2

MT 44

Solutions

1.

,

So remainder = 3

83 = 512

Unit digit = 2

2. 248.6+25.6+28.2

= (16)25.55

3. 343 – 9 = unit digit = 4

4.

Remainder is 3.

5.

6.

So, No. of zero = 84 + 16 + 3 = 103

7. = 13p + 11 and = 17q + 9

13p + 11 = 17q + 9

q =

The least value of P for which q = 2 + 13p, a whole number as p = 26

= 13 x 26 + 11 = 349

8.

Remainder = 4 – 1 = 3

9. 3 × 6 × 3 = 4 [unit digit]

10. mn = (11)

2 = 121

So, m = 11, n = 2

(m – 1)n+1

= 1000

11.

MT 45

Solutions

1. 2M + 1T = 7000

2T + 1M = 9800

By, elimination; 1T = 4200

2. 24x = 18 (x + 2)

x = 6 to each person

3. (8P + 6P = 100) × 1.5 = 12P + 9P = 150

4. (4 × 3) + (4.5 × x) = 39 → x = 6 hours

Total hours = 4 + 6 = 10 hours

5. H + C = 48

2H + 4C = 140

By elimination, H = 26

6. 10C = 4T; so, 5C = 2T

15C + 2T = 40

15C + 5C = 40 [replace 2T = 5C]

C = 2, T = 5

7. L.C.M. of 3, 2 = 6 min

In every 6 min , A will type = 4 pages

B will type = 9 pages

i.e. A+B = 13 pages

So, time taken =

min

8. Average =

so greatest No. = 22 [14 is middle no.]

9. First No. = x

Second No. = 2x (In Ratio of 1 : 2)

so, 9 (x – y) = 36

9 (2x – x) = 36

x = 4

2x = 8

So, difference between sum and difference of digits = 12 – 4 = 8

10. a2 + b

2 + c

2 = 138 and ab + bc + ca = 131

= a2 + b

2 + c

2 + 2(ab + bc + ca)

= 138 + 2 x 131

So, a + b + c = 20.

MT 46

Solutions

1. Son’s present age = x

So, 38 – x = x; so x = 19

5 years back son’s age = 19 – 5 = 14 years

2. Suppose, C’s age = x

then, x + 2x + (2x + 2) = 27 →x = 5; B’s age = 10 years

3.

3 years hence,

so, x = 6

Anand’s age = 4x = 24 years

4. 10 years ago, son = x years; then, father = 3x

10 years later, son = x + 20

father = 3x + 20

so, 3x + 20 = 2 (x + 20); x = 20

5.

6. 5 years ago; son = x years

father = y years

Given, xy = 4 y x = 4

At present (x + 5) + (y + 5) = 45

x + y = 35

So, y = 31 years

Present age of father = 31 + 5 = 36 years

7.

so, x = 14 ;

Younger = x = 14 years; Elder = x + 16 = 30 years

8. (x + 10) + (2x + 10) + (3x + 10) = 90

B’s age = 2x + 10 = 40 years

9. Mother = 5x

Person = 2x

So,

; So, x = 8; Mother’s age = 40

10. If Kamala = x years

x =

so, x = 30

son’s age =

years

MT 47

Solutions

1. A → 6 days

B → 8 days

C → 12 days

Distribution of money in ratio of

i.e. C’s share =

2. Efficiency Ratio = 2 : 1

Time (days) Ratio = 1 : 2

In 10 days they will do (2 + 1) × 10 = 30 units

So, Paul is more (2 times) efficient, so takes 15 days

and Paul takes 30 days

3. Work done by Ishan in 10 days =

work

Remaining

work done by Bharat in 4 days

Bharat takes 9 days to complete whole work together

work; so, 6 days

4. 1 M → 100 days 1 W → 120 days 1 C → 200 days

2 M → 50 days 3 W → 40 days 2 C →100 days

So working together by 2M + 3W + 2C =

days

5.

work

Remaining

done by C; so time taken = 4 days

6. 24W = 20M 40B = 20M

6 W = 5M 2B = 1M

So,

→ x = 128 – 6 = 122 Men

7. 4M + 3W =

5M + 6M =

So, 1W = 54 days (by elimination)

8. In first 2 hours, A and B sow

of field

work done in 4 laps; here 4 = 4 × 2 = 8 hours

Remaining

work, first

will be done by ‘A’ in 1 hour, and rest

by win B in

an

hour.

So, total time =

hours.

9. 30 × 7 × 18 = 21 × 8 × D2

D2 =

days

10.

So,

days

MT 48

Solutions

1.

Work done in 2 minutes

Remaining

work by second tap in 13 minutes

2. Together =

of tank

So,

hours = 5.45 hours

3.

→ x = 12 hours

4.

work in 10 minutes

So, remaining

by another tap in 20 minutes.

5. Speed of empty tap in 2 times more than filling pipe. So, tank will be empty, if both pipes open.

6. In 9 minutes A will do

Remaining

work will be done by B =

minutes

7. Time

so, 6 hours

8. Together =

so, 5 minutes

9. Together

hours

10.

in first 2 hours.

work done in 2 laps and here

2 = 2 × 2 = 4 hours

Remaining work =

work, that will be done by ‘A’ in next 1 hour.

So, total time = 4 + 1 = 5 hours.

MT 49

Solutions

1.

kmph

2.

kmph

3.

kms

4. 11 – 9 = 2

If 2 part = 60 m

11 part = ? = 330 m

5.

L = 400 m

6. Train covers its length in 18 sec.

So, platform’s length in rest 12 sec.

Length = 75 ×

× 12 = 250 m

7.

kms

8.

So, S1 = 87.5 kmph

9. 14 – 10 = 4 kmph more speed, so walked 20 km more

4 → 20

10 → ? = 50 kms

10. Avg. speed =

kmph

MT 50

Solutions

1. Upstream = 20 – 4 = 16 kmph

Downstream = 20 + 24 = 24 kmph

hrs.

So, upstream in 5 hours = 16 × 5 = 80 kms

2. Speed of boat =

kmph

Speed of stream =

kmph

3. Velocity of stream =

kmph

4. Downstream = 10 + 6 = 16 kmph

Time taken =

hours

5. Upstream = B = 3 ;

kmph

6. Upstream = 5 – 1 = 4 kmph

Downstream = 5 + 1 = 6 kmph

Distance =

kms

7. Downstream = 13 + 4 = 17 kmph

Time taken =

hours

8. Man’s speed =

kmph

Stream’s speed =

kmph

9. Stream’s speed =

kmph

10. Downstream = 15 + 3 = 18 kmph

Upstream = 15 – 3 = 12 kmph

Ratio =