hints and solutions to mathematics - ique ideas 35-50 hints a… · · 2016-09-26... husband’s...
TRANSCRIPT
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.
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 =