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A new symbol can define a custom operation · 10 practice problems

4.OA.A.35.OA.A.1

0 / 10 solved

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+20a \diamond b = a \times b + 20. Find the value of (24⋄13)⋄17(24 \diamond 13) \diamond 17.

Variant 2 easy

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+30a \diamond b = a \times b + 30. Find the value of (32⋄15)⋄23(32 \diamond 15) \diamond 23.

Variant 3 medium

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+35a \diamond b = a \times b + 35. Find the value of (29⋄14)⋄18(29 \diamond 14) \diamond 18.

Variant 4 easy

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+10a \diamond b = a \times b + 10. Find the value of (35⋄16)⋄21(35 \diamond 16) \diamond 21.

Variant 5 medium

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+40a \diamond b = a \times b + 40. Find the value of (18⋄22)⋄14(18 \diamond 22) \diamond 14.

Variant 6 medium

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+25a \diamond b = a \times b + 25. Find the value of (41⋄12)⋄19(41 \diamond 12) \diamond 19.

Variant 7 medium

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+15a \diamond b = a \times b + 15. Find the value of (46⋄11)⋄15(46 \diamond 11) \diamond 15.

Variant 8 hard

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+50a \diamond b = a \times b + 50. Find the value of (27⋄19)⋄13(27 \diamond 19) \diamond 13.

Variant 9 hard

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+45a \diamond b = a \times b + 45. Find the value of (52⋄13)⋄12(52 \diamond 13) \diamond 12.

Variant 10 hard

For any two numbers aa and bb, the operation ⋄\diamond is defined by a⋄b=a×b+60a \diamond b = a \times b + 60. Find the value of (38⋄17)⋄16(38 \diamond 17) \diamond 16.

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