Let Us Work Out – Simplification Problems
Mathematics becomes easier when we learn to simplify expressions step by step. In these exercises from Ganit Prabha Class 6, we will practice the BODMAS rule, handle brackets, and solve real-life story sums.
BODMAS stands for:
B – Brackets (solve inside brackets first)
O – Orders (powers, roots, etc.)
D – Division (from left to right)
M – Multiplication (from left to right)
A – Addition (from left to right)
S – Subtraction (from left to right)
This rule gives the correct order of operations. Without BODMAS, answers can be wrong.
1.(A) Verify, if the values are same for all these cases
(a) 20 + 8 ÷ (4 − 2)
Solution:
20 + 8 ÷ (4 − 2)
= 20 + 8 ÷ 2
= 20 + 4
= 24
(b) (20 + 8) ÷ (4 − 2)
Solution:
(20 + 8) ÷ (4 − 2)
= 28 ÷ 2
= 14
(c) (20 − 8)(4 − 2)
Solution:
(20 − 8)(4 − 2)
= 12 × 2
= 24
(d) 20 − 8 (4 − 2)
Solution:
= 20 − 8 × 2
= 20 − 16
= 4
(e) (20 + 8) ÷ 4 − 2
Solution:
(20 + 8) ÷ 4 − 2
= 28 ÷ 4 − 2
= 7 − 2
= 5
1.(B) Let us form similar simplification sums with numbers 12, 6, 3 and 1 and then find their values.
(a) 12 + 6 ÷ (3 − 1)
Solution:
= 12 + 6 ÷ 2
= 12 + 3
= 15
(b) (12 + 6) ÷ (3 − 1)
Solution:
= 18 ÷ 2
= 9
(c) (12 − 6)(3 − 1)
Solution:
= 6 × 2
= 12
(d) 12 − 6 (3 − 1)
Solution:
= 12 − 6 × 2
= 12 − 12
= 0
(e) (12 + 6) ÷ 3 − 1
Solution:
= 18 ÷ 3 − 1
= 6 − 1
= 5
2. Let us find the value of simplification sums:
(a) \(256 \div \overline {16 \div 2} \div \overline {18 \div 9} \times 2\)
Solution:
\(256 \div \overline {16 \div 2} \div \overline {18 \div 9} \times 2\)
= 256 ÷ (16 ÷ 2) ÷ (18 ÷ 9) × 2
= 256 ÷ 8 ÷ 2 × 2
= 32 ÷ 2 × 2
= 16 × 2
= 32
(b) (72 ÷ 8 × 9) − (72 ÷ 8 of 9)
Solution:
(72 ÷ 8 × 9) − (72 ÷ 8 of 9)
= (9 × 9) − (9 × 9)
= 81 − 81
= 0
(c) \(76 - 4 - \left[ {6 + \{ 19 - \left( {48 - \overline {57 - 17} } \right)\} } \right]\)
Solution:
\(76 - 4 - \left[ {6 + \{ 19 - \left( {48 - \overline {57 - 17} } \right)\} } \right]\)
= 76 − 4 − [6 + {19 − (48 − (57 − 17))}]
= 76 − 4 − [6 + {19 − (48 − 40)}]
= 76 − 4 − [6 + {19 − 8}]
= 76 − 4 − [6 + 11]
= 76 − 4 − 17
= 72 − 17
= 55
(d) {25 × 16 ÷ (60 ÷ 15) − 4 × (77 − 62)} ÷ (20 × 6 ÷ 3)
Solution:
{25 × 16 ÷ (60 ÷ 15) − 4 × (77 − 62)} ÷ (20 × 6 ÷ 3)
= {400 ÷ 4 − 4 × 15} ÷ (120 ÷ 3)
= {100 − 60} ÷ 40
= 40 ÷ 40
= 1
(e) \(\left[ {16 \div \left( {42 - \overline {38 + 2} } \right)} \right] \times 12 \div \left( {24 \div 6} \right) \times 2 + 4\)
Solution:
\(\left[ {16 \div \left( {42 - \overline {38 + 2} } \right)} \right] \times 12 \div \left( {24 \div 6} \right) \times 2 + 4\)
= [16 ÷ (42 − (38 + 2))] × 12 ÷ (24 ÷ 6) × 2 + 4
= [16 ÷ (42 − 40)] × 12 ÷ 4 × 2 + 4
= [16 ÷ 2] × 12 ÷ 4 × 2 + 4
= 8 × 12 ÷ 4 × 2 + 4
= 96 ÷ 4 × 2 + 4
= 24 × 2 + 4
= 48 + 4
= 52
(f) \(4 \times \left[ {24 - \{ \left( {110 - \overline {11{\text{ }} + {\text{ }}3} {\text{ }} \times {\text{ }}4} \right) \div {\text{ }}9\} } \right] \div 2\;{\text{of }}9\)
Solution:
\(4 \times \left[ {24 - \{ \left( {110 - \overline {11{\text{ }} + {\text{ }}3} {\text{ }} \times {\text{ }}4} \right) \div {\text{ }}9\} } \right] \div 2\;{\text{of }}9\)
= 4 × [24 − {(110 − 14 × 4) ÷ 9}] ÷ 2 × 9
= 4 × [24 − {(110 − 56) ÷ 9}] ÷ 2 × 9
= 4 × [24 − (54 ÷ 9)] ÷ 2 × 9
= 4 × [24 − 6] ÷ 2 × 9
= 4 × 18 ÷ 2 × 9
= 72 ÷ 2 × 9
= 36 × 9
= 324
(g) 200 ÷ [88 − {(12 × 13) − 3 × (40 − 9)}]
Solution:
200 ÷ [88 − {(12 × 13) − 3 × (40 − 9)}]
= 200 ÷ [88 − (156 − 93)]
= 200 ÷ [88 − 63]
= 200 ÷ 25
= 8
(h) \[\left( {987 - \overline {43{\text{ }} + {\text{ }}25} } \right) - 10\left[ {5 + \{ \left( {999 \div \overline {9 \times 3} } \right) + \left( {\overline {8 \times 9} \div 6} \right){\text{ }} \times 4\} } \right]\]
Solution:
\[\left( {987 - \overline {43{\text{ }} + {\text{ }}25} } \right) - 10\left[ {5 + \{ \left( {999 \div \overline {9 \times 3} } \right) + \left( {\overline {8 \times 9} \div 6} \right){\text{ }} \times 4\} } \right]\]
= (987 − (43 + 25)) − 10 [5 + {(999 ÷ (9 × 3)) + ((8 × 9) ÷ 6) × 4}]
= (987 − 68) − 10 [5 + {(999 ÷ 27) + (72 ÷ 6) × 4}]
= 919 − 10 [5 + {37 + 12 × 4}]
= 919 − 10 [5 + {37 + 48}]
= 919 − 10 [5 + 85]
= 919 − 10 × 90
= 919 − 900
= 19
3. Let us form a story for following simplification and then solve.
(a) (12 − 2) ÷ 2
Solution:
Story: There were 12 apples. 2 apples were eaten. The remaining apples are shared equally between 2 children. How many apples does each child get?
Therefore, (12 − 2) ÷ 2
= 10 ÷ 2
= 5
Answer: Each child will get 5 apples.
(b) {90 − (48 − 21)} ÷ 7
Solution:
Story: A farmer harvested 90 mangoes. He gave 48 mangoes to his neighbor but got back 21 from him. The farmer then divided the remaining mangoes equally among 7 baskets. How many mangoes were in each basket?
Therefore, {90 − (48 − 21)} ÷ 7
= (90 − 27) ÷ 7
= 63 ÷ 7
= 9
Answer: 9 mangoes were in each basket.
4. Let us express in mathematical language and solve.
Solution:
Money from selling guavas = 125 × 2 = Rs 250
Cost of 2 pens = 2 × 5 = Rs 10
Cost of 2 exercise books = 2 × 20 = Rs 40
Total spent = 10 + 40 = Rs 50
Remaining money = 250 − 50 = Rs 200
Money shared among 4 children = 200 ÷ 4 = Rs 50
Answer: Rajdeep got Rs 50 for buying sweets.
Hi Please, do not Spam in Comments.