# Free Fractions and Decimals 02 Practice Test - 7th grade

Solve 13(56×918)+29.

A.

436

B.

536

C.

636

D.

736

#### SOLUTION

Solution : B

13(56×918)+29

=13(1536)+29

=131536+29

Taking L.C.M of denominators, 1215+836=536

12×(1+14)×(1+24)×(1+34)×(1+54) is equal to:

A.

44256

B.

948255

C.

1890512

D.

945256

#### SOLUTION

Solution : C and D

12×(1+14)×(1+24)×(1+34)×(1+54)

= 12 ×54×64×74×94=945256

Find the least fraction, which must be multiplied to (124 12 +3616 116 +28) to make it an integer.

A.

241

B.

161

C.

731

D.

173

#### SOLUTION

Solution : B

(124 -12 +3616 -116 +28)

=64 12 +5416 116 +28

Taking LCM of denominators and solving,

24 8 +54 1 +416=7316

Now, If we multiply 7316 by 161, it becomes 73, which is an integer

Hence, 161 is the required fraction.

The value of (12)×(12)+(12)(21)×(12)×(34) is ___.

#### SOLUTION

Solution :

(12)×(12)+(12)(12)+(12)×(34) = (14)+(12)1×(34) = (34)(34)=1

[51425×559+(12+34)] is equal to:

A.

916

B.

7718

C.

2412

D.

1224

#### SOLUTION

Solution : B

Given [51425×559+(12+34)]

Apply BODMAS rule to solve

= [214(25×509)+54]

= [214(209)+54]

Taking LCM of denominators, we have

18980+4536=15436=7718

[51425×559+(12+34)]=7718

A person bought 312 kg of apples, 223 kg of grapes and 113 kg of oranges. He wants to divide them among 4 people. Find the amount of fruits(in kg) that each person will receive.

A.

138 kg

B.

158 kg

C.

178 kg

D.

278 kg

#### SOLUTION

Solution : C

First, let us do the conversion of mixed fractions into improper fractions.

i.e  312=(2×3)+12=72

223=(3×2)+23=83, and

113=(3×1)+13=43

Total quantity of fruits=(72+83+43) kg

=72+4

=72+82 kg

=152 kg

When 152 kg is divided among 4 people,

each person's share =1524kg

=(152×14)kg

=158 kg

=178 kg

Hence, each person gets (178) kg of fruits.

212+213+214110×25×1256 is equals to:

A.

5.5

B.

6.5

C.

7.5

D.

8.5

#### SOLUTION

Solution : D

52+73+94110×25×1256

=851256

=8512×65

=172=8.5

Varun travelled 18​th of the journey which is equal to 56km on a scooter . What is the remaining distance he has to cover?

A.

342 km

B.

392 km

C.

352 km

D.

442 km

#### SOLUTION

Solution : B

Let the total distance to be covered =d km

Then, 18×d = 56km

d = (56×8)km = 448km

Remaining distance = (448 - 56)km = 392km

The product of two fractions is 20. If one of the fractions is 14, then the other fraction is ___.

#### SOLUTION

Solution : Let the other fraction be x

Then, x×14=20

x=20×4=80

Simplify the following expression and choose the correct answer:

0×(59×23)+56×9×0(45+35+12)

A.

910

B.

49

C.

1

D.

0

#### SOLUTION

Solution : D

Concentrate on numerator. The expression has '0' multiplied to it, and any number multiplied by '0' is '0'.

And 0 divided by any number always '0'.