Topic - Quant Mixed Bag

Solved ? - Yes

Source - ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - ]]>

a. 169.5

b. 170

c.175

d.180. ]]>

= > 126 + 135 + 2t = 612

= > 261 + 2t = 612

= > t = 175.50.

Short cut :

The average of mixture here is 153.We take it as base.

1st tea rate is -27 from base.2nd tea rate is -18 from base.

So obviously rate of 3rd tea must be higher from the base.

So -27-18+2t = 0

t = +22.5.

so rate of third tea = 153 + 22.5 = 175.50.

a. 78

b. 80

c. 91

d. 94 ]]>

= 17 x 78 – 16 x 77

= (16+1)(77+1) – 16x77

= 16x77+16+77+1-16x77

= 94.

Short approach :-

Previous average(16 students) = 77.

New average(17 students) = 78.

While 17th student appeared for exam,to maintain average of 77 he had to get atleast 77 marks.

(But new average is 78,it is only possible when each student will get +1 mark)

Marks of 17th student = 77 + 1 x 17 = 94.

a. 99

b. 149

c. 151

d. CBD ]]>

After 49 innings score = 2304+97=2401.

Average after 49 innings = 2401/49=49.

Required score = (49+2)x50 – 2401 = 149.

Short approach :-

Average till 48 inning = 48.

In 49th inning he scored 97 runs = 48+49.

New average = (48+49)+48x48/49 = 49.

Required score in 50 to increase average by 2 = 49+2x50 = 149

a. 17 kg

b. 20 kg

c. 36 kg

d. 31 kg ]]>

A + B = 2 x 40 = 80.

B + C=2 x 43 = 86.

C = 135 – 80 = 55.

A = 135 – 86 = 49.

B = 135 – 49 – 55 = 31.

Short approach :-

Average of A,B,C = 45. (base)

Average of A & B = -5 (from base)

Average of B & C = -2 (from base)

Weight of B = 45 – 2x5 – 2x2 = 31.

a. 4991

b. 5991

c. 6001

d. 6991. ]]>

6435+6927+6855+7230+6562+z = 6 x 6500

= > 34009 + z = 39000

= > z = 4991.

Short approach :-

let's take 6500 as base.

6500-(-65+427+355+730+62)

=6500 - (1509)

= 4991.

no of boys = a+6

47(a+6)+42a = 44.8(a+a+6)

47a+42a+282 = 44.8(2a+6)

89a + 282 = 89.6a + 268.8

0.6 a = 13.2

a = 22.

a+6 = 28.

total = 22+28 = 50.

Short approach :-

average of all class=44.8.

ratio of boys & girls= (47-44.8 ):(44.8-42)=2.2:2.8.

(2.8-2.2)x=6

x=10.

total students= (2.2+2.8 )x10=50

weight of new player = k.

11a - 55 + k = 11(a+1)

k - 55 = 11

k = 66.

Short approach :-

weight of new player = 55 + 1 x 11 = 66.

= > 60th value = 71 x 48 - 59 x 46 - 11 x 52.

= > 60th value = 122.

Short approach :-

We take 48 as base.

59 results average is -2 (from base)

11 results average is +4 (from base)

So 59 x (-2) + 11 x 4 = -74.

To get 60th value this value should be decrease from base.

Hence 60th value = 48 - (-74) = 122.

two numbers discarded (45,55)

new total = 1900-45-55 = 1800.

new average = 1800 / 48 = 37.5

Short approach :-

The average of 50 numbers is 38.

so if we discarded two number then to maintain average of 38,value should be decreased by 2 x 38 = 76.

but here value decreased by 100.

So 100-76 = 24.

There is 48 numbers.

so every number value will be decreased by 24/48 = 0.5

new average = 38 - 0.5 = 37.5.

Virat + Rahane = 2 x 200 = 400.

Total runs = 560 + 400 = 960.

Average = 960 / 4 = 240.

Short approach :-

Take average of any two

let we take Rohit & Dhawan,which is nothing but 280.(base)

now average of virat & rahane 200.

which is -80 (from base)

Average = 280 - (80/2)

= 280 - 40

= 240.

We can solve it by take 200 too..

base = 200.

average of Rohit & Dhawan is +80(from base).

average = 200 + (80/2)

= 240.

Here divided by 2, because that is average of two person..

new weight (including teacher) = 30 x 40.5 = 1215.

Teacher's weight = 1215 - 1160 = 55 kg.

Short approach :-

base = 40.

including teacher average increased by 0.5 kg.

Teacher's weight = 40 + 30 x 0.5

= 40 + 15

= 55 kg.

cp of 1 orange = y.

18x – 18y = 3x

= > 18 y = 15 x.

Profit percentage = (3x/18y)x100

= (3x/15x)x100

= 20%.

Short approach :-

Profit percentage = [profit (quantity) / (Total quantity – profit)] x 100

= [(3)/(18-3)] x 100

= [3/15] x 100

= 20%