Free Decimals 01 Practice Test - 4th Grade
Question 1
18 as a decimal =
125
12.5
1.25
0.125
SOLUTION
Solution : D
8's friend is 125 as 8×125=1000
18×125125=1251000=0.125
Question 2
Find the value.
500+4+4100=___
SOLUTION
Solution : D
We have to do the addition:
=500+4+4100
=500+4+0.04
5 0 0 . 00 4 . 00+ 0 . 04–––––––––– 5 0 4 . 04––––––––––
Question 3
If one pencil costs ₹ 5.75, then the cost of 100 such pencils is _____.
₹ 0.575
₹ 57.50
₹ 575
₹ 5750
SOLUTION
Solution : C
Given: Cost of 1 pencil = ₹ 5.75
∴ Cost of 100 pencils =₹ (5.75×100)
⇒ Cost of 100 pencils = ₹ 575
Question 4
1.1−0.03= ____
1.7
1.77
1.07
1.70
SOLUTION
Solution : C
Since 0.03 has two decimal places, we use the equivalent decimal of 1.1 with two decimal places.
i.e., 1.1−0.03=1.10−0.0.3
1 ./10/010− 0 . 0 3––––––––––– 1 . 0 7––––––––––
Question 5
A shopkeeper wants to distribute 20.7 kgs of sugar into 9 equal jars. How much sugar would go into each jar?
2.7 kg
2.1 kg
2.3 kg
3.1 kg
SOLUTION
Solution : C
To find the amount of sugar that goes into each jar, we need to divide 20.7 kg by 9.
∴20.79 kg=2.3 kg∴ Each jar will have 2.3 kg of sugar.
Question 6
Find the value after shifting the decimals.
4.5×1.5 = ____
3.75
4.75
6.75
2.57
SOLUTION
Solution : C
We first find 45×15, which is equal to 675.0
Shifting the decimal point two places to the left of 675.0, we get 6.75.
Therefore, 4.5×1.5=6.75.
Question 7
Six cases of paper cost ₹ 156.49. How much does one cost?
₹ .26.19
₹ 25.45
₹ 26.08
₹ 25.18
SOLUTION
Solution : C
cost of six cases = ₹ 156.49
∴ Cost of one case =156.496
156.496=26.08166...
Approximately equal to ₹ 26.08 (Rounding off).
Question 8
Compare the following and fill in with an appropriate symbol
>
<
=
Not comparable
SOLUTION
Solution : C
In the first figure, 27 parts are shaded out of 100 , so the decimal is 0.27
In the second figure, 27 parts are shaded out of 100, so the decimal is 0.27Hence both the decimals are equal
Question 9
3×46+4×512−5×710 =
56
13
16
35
SOLUTION
Solution : C
3×46+4×512−5×7103×46+4×512−5×710∴42+53−72∴LCM of 2 ×3=6∴4×32×3+5×23×2−7×32×312+10−2122−216=16
Question 10
Find : 3÷67+7÷212 −2÷5
31330
32330
32130
31130
SOLUTION
Solution : B
3÷67+7÷212−2÷5∴3×76+7×221−25∴72+23−25LCM of 2, 3, 5 is 30∴72=7×152×15=1053023=2×103×10=203025=2×65×6=1230∴105+20−1230=11330∴11330=32330